New Bounds for the Euclidean Numerical Radius of Two Operators in Hilbert Spaces

Abstract

This paper presents new weighted lower and upper bounds for the Euclidean numerical radius of pairs of operators in Hilbert spaces. We show that some of these bounds improve on recent results in the literature. We also find new inequalities for the numerical radius and the Davis–Wielandt radius. The lower and upper bounds we obtain are not symmetrical.

1. Introduction

Let H be a complex Hilbert space with inner product $⟨·,·⟩$ and corresponding norm $\parallel ·\parallel$. We represent the $C^{\ast }$-algebra of all bounded linear operators on H by $\mathcal{L}\left(H\right)$. For any operator $A\in \mathcal{L}\left(H\right)$, we denote its adjoint by $A^{\ast }$, and let $\left|A\right|={\left(A^{\ast }A\right)}^{1/2}$ indicate the positive square root of $A^{\ast }A$. The real and imaginary parts of A are defined as $\mathfrak{R}\left(A\right)={\displaystyle \frac{1}{2}}(A+A^{\ast })$ and $\mathfrak{I}\left(A\right)={\displaystyle \frac{1}{2\mathrm{i}}}(A-A^{\ast })$, respectively. The numerical range of A, written as $W\left(A\right)$, is the set $\{⟨Ax,x⟩:x\in H,\parallel x\parallel =1\}$. For simplicity, we use “operator” to refer to an operator in $\mathcal{L}\left(H\right)$.

Let $\parallel A\parallel$ and $w\left(A\right)$ represent the operator norm and numerical radius of the operator A, respectively. The operator norm of A is defined as

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

and the numerical radius of A is given by

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

It is established that the numerical radius $w(·)$ defines a norm on $\mathcal{L}\left(H\right)$ and is comparable to the operator norm $\parallel ·\parallel$. Specifically, the following double inequality holds:

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

(1)

These inequalities are sharp. The first inequality attains equality if $A^2=0$, while the second inequality attains equality if A is a normal operator, i.e., $A^{\ast }A=A{A}^{\ast }$. Kittaneh [1] presented a refinement of these inequalities by establishing that

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

(2)

For further developments on (1) and (2), readers may refer to [2,3,4,5,6,7,8] and the references therein.

The Davis–Wielandt radius of an operator $T\in \mathcal{L}\left(H\right)$, denoted by $dw\left(T\right)$, is defined in [9,10] as

$$dw\left(T\right)=\underset{\parallel x\parallel =1}{sup}{\left({\left|⟨Tx,x⟩\right|}^2+{\parallel Tx\parallel }^4\right)}^{1/2}.$$

It follows that $dw\left(T\right)\ge 0$ and $dw\left(T\right)=0$ if, and only if, $T=0$. For any scalar $\eta \in \mathbb{C}$, the Davis–Wielandt radius satisfies $dw\left(\eta T\right)>\left|\eta \right|dw\left(T\right)$ if $\left|\eta \right|>1$, $dw\left(\eta T\right)<\left|\eta \right|dw\left(T\right)$ if $\left|\eta \right|<1$, and $dw\left(\eta T\right)=\left|\eta \right|dw\left(T\right)$ when $\left|\eta \right|=1$.

Note that the triangle inequality $dw(T+S)\le dw\left(T\right)+dw\left(S\right)$ does not hold for arbitrary $T,S\in \mathcal{L}\left(H\right)$. However, this inequality is valid when $\mathfrak{R}\left(T^{\ast }S\right)=0$ (see ([11], Corollary 2.2)). Additionally, it is straightforward to show that

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

and these inequalities are sharp (see ([11], Corollary 2.2)).

Zamani and Shebrawi ([12], Theorem 2.1) established that

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

Additionally, in ([12], Theorems 2.13, 2.14, and 2.17), the authors showed that

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

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

and

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

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

In recent work, Bhunia et al. ([11], Theorem 2.4) provided the following upper bound:

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

(3)

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

Additionally, in [13], the authors derived inequalities for the sum of operators, as follows:

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

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

Let $(C,D)$ be a pair of bounded linear operators on H. The Euclidean operator radius is defined as:

$$w_e(C,D):=\underset{\parallel x\parallel =1}{sup}{\left({\left|⟨Cx,x⟩\right|}^2+{\left|⟨Dx,x⟩\right|}^2\right)}^{1/2}.$$

As noted in [14], $w_e:{\mathcal{B}}^2\left(H\right):=\mathcal{B}\left(H\right)\times \mathcal{B}\left(H\right)\to [0,\infty )$ defines a norm, and the following inequality is satisfied:

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

(4)

for $(C,D)\in {\mathcal{B}}^2\left(H\right)$, where the constants ${\displaystyle \frac{\sqrt{2}}{4}}$ and 1 are optimal in (4). For more details on the Euclidean operator radius, related results, and their generalizations, see [15,16,17,18,19,20] and the references cited therein.

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

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

By setting $C=T$ and $D={\left|T\right|}^2$ in (4), we obtain

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

which provides the upper bound from (3) and a similar lower bound.

Motivated by the above results, in this paper, we present some new weighted lower and upper bounds for the Euclidean numerical radius of a pair of operators and show that some of them are sharper than certain recent results obtained by other authors. As a natural consequence, we also derive new inequalities for the numerical radius and the Davis–Wielandt radius.

Among other results, we obtain the following lower bounds for the Euclidean numerical radius:

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

and

$$\begin{array}{cc}\hfill w_e^2\left(A,B\right)& \ge \left|\alpha \beta \right|w\left(AB+BA\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left[\alpha \left(1+i\right)A+\beta \left(1-i\right)B\right],w\left[\alpha \left(1-i\right)A+\beta \left(1+i\right)B\right]\right\}\hfill \\ \hfill & \times \left|w\left(\alpha A+\beta B\right)-w\left(\alpha A-\beta B\right)\right|,\hfill \end{array}$$

for $A,B\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2=1.$

Let f, g, h, and k be nonnegative continuous functions on $[0,\infty )$ satisfying the condition $f\left(t\right)g\left(t\right)=h\left(t\right)k\left(t\right)=t$ for all $t\in [0,\infty )$. If $A,B\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta ,\gamma ,\delta \in \{-1,1\}$, then we also obtain the following upper bounds:

$$w_e^2\left(A,B\right)\le {∥f^{2p}\left(\left|A\right|\right)+h^{2p}\left(\left|B\right|\right)∥}^{1/p}{∥g^{2q}\left(\left|A^{\ast }\right|\right)+k^{2q}\left(\left|B^{\ast }\right|\right)∥}^{1/q},$$

and

$$w_e^2\left(A,B\right)\le {∥f^{2p}\left(\left|A\right|\right)+k^{2p}\left(\left|B^{\ast }\right|\right)∥}^{1/p}{∥g^{2q}\left(\left|A^{\ast }\right|\right)+h^{2q}\left(\left|B\right|\right)∥}^{1/q},$$

for $A,B\in \mathcal{L}\left(H\right)$ and $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1.$

Applications for the numerical radius and the Davis–Wielandt radius are also provided.

2. Main Results

In this section, we present our main results, beginning with the following initial result.

Theorem 1.

For $A,B\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta ,\gamma ,\delta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2={\left|\gamma \right|}^2+{\left|\delta \right|}^2=1,$ we have

$$\begin{array}{cc}\hfill & w_e^2\left(A,B\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}max\left\{w\left[\left({\alpha }^2+{\gamma }^2\right)A^2+\left({\beta }^2+{\delta }^2\right)B^2+\left(\alpha \beta +\gamma \delta \right)\left(AB+BA\right)\right],\right.\hfill \\ \hfill & \left.w\left[\left({\alpha }^2+{\gamma }^2\right)A^2+\left({\beta }^2+{\delta }^2\right)B^2-\left(\alpha \beta +\gamma \delta \right)\left(AB+BA\right)\right]\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left[\left(\alpha +\gamma \right)A+\left(\beta +\delta \right)B\right],w\left[\left(\alpha -\gamma \right)A+\left(\beta -\delta \right)B\right]\right\}\hfill \\ \hfill & \times \left|w\left(\alpha A+\beta B\right)-w\left(\gamma A+\delta B\right)\right|,\hfill \end{array}$$

(5)

and

$$\begin{array}{cc}\hfill & w_e^2\left(A,B\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{4}}max\left\{w\left[\left[{\left(\alpha +\beta \right)}^2+{\left(\gamma +\delta \right)}^2\right]A^2+\left[{\left(\alpha -\beta \right)}^2+{\left(\gamma -\delta \right)}^2\right]B^2\right.\right.\hfill \\ \hfill & +\left({\alpha }^2+{\gamma }^2-{\beta }^2-{\delta }^2\right)\left(AB+BA\right),\hfill \\ \hfill & w\left[\left[{\left(\alpha -\beta \right)}^2+{\left(\gamma -\delta \right)}^2\right]A^2+\left[{\left(\alpha +\beta \right)}^2+{\left(\gamma +\delta \right)}^2\right]B^2\right.\hfill \\ \hfill & \left.+\left({\alpha }^2+{\gamma }^2-{\beta }^2-{\delta }^2\right)\left(AB+BA\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}max\left\{w\left[\left(\alpha +\gamma +\beta +\delta \right)A+\left(\alpha +\gamma -\beta -\delta \right)B\right],\right.\hfill \\ \hfill & \left.w\left[\left(\alpha +\beta -\gamma -\delta \right)A+\left(\alpha -\beta -\gamma +\delta \right)B\right]\right\}\hfill \\ \hfill & \times \left|w\left(\left(\alpha +\beta \right)A+\left(\alpha -\beta \right)B\right)-w\left(\left(\gamma +\delta \right)A+\left(\gamma -\delta \right)B\right)\right|.\hfill \end{array}$$

(6)

Proof. 

By using the Cauchy–Schwarz inequality for ${\left|\alpha \right|}^2+{\left|\beta \right|}^2=1,$ we have

$$\begin{array}{ccc}\hfill {\left|⟨Ax,x⟩\right|}^2+{\left|⟨Bx,x⟩\right|}^2& =& \left({\left|\alpha \right|}^2+{\left|\beta \right|}^2\right)\left({\left|⟨Ax,x⟩\right|}^2+{\left|⟨Bx,x⟩\right|}^2\right)\hfill \\ & \ge & {\left(\left|\alpha \right|\left|⟨Ax,x⟩\right|+\left|\beta \right|\left|⟨Bx,x⟩\right|\right)}^2\hfill \\ & =& {\left(\left|⟨\alpha Ax,x⟩\right|+\left|⟨\beta Bx,x⟩\right|\right)}^2\hfill \\ & \ge & {\left|⟨\alpha Ax,x⟩+⟨\beta Bx,x⟩\right|}^2={\left|⟨\left(\alpha A+\beta B\right)x,x⟩\right|}^2,\hfill \end{array}$$

for all $x\in H.$

If we take the supremum over $x\in H,$$∥x∥=1$, we obtain

$$w_e^2\left(A,B\right)\ge w^2\left(\alpha A+\beta B\right).$$

Also, if ${\left|\gamma \right|}^2+{\left|\delta \right|}^2=1,$ we have

$$w_e^2\left(A,B\right)\ge w^2\left(\gamma A+\delta B\right).$$

Therefore,

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

(7)

Now, observe that

$$\begin{array}{cc}\hfill & w^2\left(\alpha A+\beta B\right)+w^2\left(\gamma A+\delta B\right)\hfill \\ \hfill & \ge w\left[{\left(\alpha A+\beta B\right)}^2\right]+w\left[{\left(\gamma A+\delta B\right)}^2\right]\hfill \\ \hfill & \ge max\left\{w\left[{\left(\alpha A+\beta B\right)}^2+{\left(\gamma A+\delta B\right)}^2\right],w\left[{\left(\alpha A+\beta B\right)}^2-{\left(\gamma A+\delta B\right)}^2\right]\right\}\hfill \\ \hfill & =max\left\{w\left[\left({\alpha }^2+{\gamma }^2\right)A^2+\left(\alpha \beta +\gamma \delta \right)\left(AB+BA\right)+\left({\beta }^2+{\delta }^2\right)B^2\right],\right.\hfill \\ \hfill & \left.w\left[\left({\alpha }^2+{\gamma }^2\right)A^2-\left(\alpha \beta +\gamma \delta \right)\left(AB+BA\right)+\left({\beta }^2+{\delta }^2\right)B^2\right]\right\},\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & w\left(\alpha A+\beta B\right)+w\left(\gamma A+\delta B\right)\hfill \\ \hfill & \ge max\left\{w\left[\left(\alpha +\gamma \right)A+\left(\beta +\delta \right)B\right],w\left[\left(\alpha -\gamma \right)A+\left(\beta -\delta \right)B\right]\right\},\hfill \end{array}$$

and by (7), we derive (5).

Now, if we replace A with $C+D$ and B with $C-D,$ then we obtain

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

(8)

Observe that

$$\begin{array}{cc}\hfill & \left({\alpha }^2+{\gamma }^2\right){\left(C+D\right)}^2+\left({\beta }^2+{\delta }^2\right){\left(C-D\right)}^2\hfill \\ \hfill & =\left({\alpha }^2+{\gamma }^2\right)\left(C^2+D^2+CD+DC\right)+\left({\beta }^2+{\delta }^2\right)\left(C^2+D^2-CD-DC\right)\hfill \\ \hfill & =\left({\alpha }^2+{\gamma }^2+{\beta }^2+{\delta }^2\right)\left(C^2+D^2\right)+\left({\alpha }^2+{\gamma }^2-{\beta }^2-{\delta }^2\right)\left(CD+DC\right)\hfill \end{array}$$

$$\begin{array}{cc}\hfill & \left(C+D\right)\left(C-D\right)+\left(C-D\right)\left(C+D\right)\hfill \\ \hfill & =C^2-D^2+DC-CD+C^2-D^2=DC+CD\hfill \\ \hfill & =2\left(C^2-D^2\right).\hfill \end{array}$$

Then,

$$\begin{array}{cc}\hfill & \left({\alpha }^2+{\gamma }^2\right){\left(C+D\right)}^2+\left({\beta }^2+{\delta }^2\right){\left(C-D\right)}^2\hfill \\ \hfill & +\left(\alpha \beta +\gamma \delta \right)\left(\left(C+D\right)\left(C-D\right)+\left(C-D\right)\left(C+D\right)\right)\hfill \\ \hfill & =\left({\alpha }^2+{\gamma }^2+{\beta }^2+{\delta }^2\right)\left(C^2+D^2\right)+\left({\alpha }^2+{\gamma }^2-{\beta }^2-{\delta }^2\right)\left(CD+DC\right)\hfill \\ \hfill & +2\left(\alpha \beta +\gamma \delta \right)\left(C^2-D^2\right)\hfill \\ \hfill & =\left({\alpha }^2+{\gamma }^2+{\beta }^2+{\delta }^2+2\left(\alpha \beta +\gamma \delta \right)\right)C^2\hfill \\ \hfill & +\left({\alpha }^2+{\gamma }^2+{\beta }^2+{\delta }^2-2\left(\alpha \beta +\gamma \delta \right)\right)D^2\hfill \\ \hfill & +\left({\alpha }^2+{\gamma }^2-{\beta }^2-{\delta }^2\right)\left(CD+DC\right)\hfill \\ \hfill & =\left[{\left(\alpha +\beta \right)}^2+{\left(\gamma +\delta \right)}^2\right]C^2+\left[{\left(\alpha -\beta \right)}^2+{\left(\gamma -\delta \right)}^2\right]D^2\hfill \\ \hfill & +\left({\alpha }^2+{\gamma }^2-{\beta }^2-{\delta }^2\right)\left(CD+DC\right),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & \left({\alpha }^2+{\gamma }^2\right){\left(C+D\right)}^2+\left({\beta }^2+{\delta }^2\right){\left(C-D\right)}^2\hfill \\ \hfill & -\left(\alpha \beta +\gamma \delta \right)\left(\left(C+D\right)\left(C-D\right)+\left(C-D\right)\left(C+D\right)\right)\hfill \\ \hfill & =\left({\alpha }^2+{\gamma }^2+{\beta }^2+{\delta }^2\right)\left(C^2+D^2\right)+\left({\alpha }^2+{\gamma }^2-{\beta }^2-{\delta }^2\right)\left(CD+DC\right)\hfill \\ \hfill & -2\left(\alpha \beta +\gamma \delta \right)\left(C^2-D^2\right)\hfill \\ \hfill & =\left({\alpha }^2+{\gamma }^2+{\beta }^2+{\delta }^2-2\left(\alpha \beta +\gamma \delta \right)\right)C^2\hfill \\ \hfill & +\left({\alpha }^2+{\gamma }^2+{\beta }^2+{\delta }^2+2\left(\alpha \beta +\gamma \delta \right)\right)D^2\hfill \\ \hfill & +\left({\alpha }^2+{\gamma }^2-{\beta }^2-{\delta }^2\right)\left(CD+DC\right)\hfill \\ \hfill & =\left[{\left(\alpha -\beta \right)}^2+{\left(\gamma -\delta \right)}^2\right]C^2+\left[{\left(\alpha +\beta \right)}^2+{\left(\gamma +\delta \right)}^2\right]D^2\hfill \\ \hfill & +\left({\alpha }^2+{\gamma }^2-{\beta }^2-{\delta }^2\right)\left(CD+DC\right).\hfill \end{array}$$

Also,

$$\left(\alpha +\gamma \right)\left(C+D\right)+\left(\beta +\delta \right)\left(C-D\right)=\left(\alpha +\gamma +\beta +\delta \right)C+\left(\alpha +\gamma -\beta -\delta \right)D,$$

and

$$\left(\alpha -\gamma \right)\left(C+D\right)+\left(\beta -\delta \right)\left(C-D\right)=\left(\alpha +\beta -\gamma -\delta \right)C+\left(\alpha -\gamma -\beta +\delta \right)D.$$

Moreover,

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

and by (8), we obtain

$$\begin{array}{cc}\hfill & 2w_e^2\left(C,D\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}max\left\{w\left[\left[{\left(\alpha +\beta \right)}^2+{\left(\gamma +\delta \right)}^2\right]C^2+\left[{\left(\alpha -\beta \right)}^2+{\left(\gamma -\delta \right)}^2\right]D^2\right.\right.\hfill \\ \hfill & +\left({\alpha }^2+{\gamma }^2-{\beta }^2-{\delta }^2\right)\left(CD+DC\right),\hfill \\ \hfill & w\left[\left[{\left(\alpha -\beta \right)}^2+{\left(\gamma -\delta \right)}^2\right]C^2+\left[{\left(\alpha +\beta \right)}^2+{\left(\gamma +\delta \right)}^2\right]D^2\right.\hfill \\ \hfill & \left.+\left({\alpha }^2+{\gamma }^2-{\beta }^2-{\delta }^2\right)\left(CD+DC\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left[\left(\alpha +\gamma +\beta +\delta \right)C+\left(\alpha +\gamma -\beta -\delta \right)D\right]\right.\hfill \\ \hfill & \left.w\left[\left(\alpha +\beta -\gamma -\delta \right)C+\left(\alpha -\gamma -\beta +\delta \right)D\right]\right\}\hfill \\ \hfill & \times \left|w\left(\left(\alpha +\beta \right)C+\left(\alpha -\beta \right)D\right)-w\left(\left(\gamma +\delta \right)C+\left(\gamma -\delta \right)D\right)\right|,\hfill \end{array}$$

and, by replacing C with A and D with $B,$ the inequality (6) is obtained. □

Remark 1.

We observe that for $\alpha =0,$$\beta =1,$$\gamma =1$ and $\delta =0$ we obtain from (5) that

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

(9)

while from (6), that

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

(10)

It should be noted that the inequality (10) was obtained earlier in ([21], Theorem 2.2). Moreover, both inequalities (9) and (10) are refinements of the inequality

$$w_e^2\left(A,B\right)\ge {\displaystyle \frac{1}{2}}w\left(A^2+B^2\right),$$

obtained in 2006 by the second author, see [22].

Remark 2.

If we take $\gamma =\alpha ,$$\delta =\beta$ in (5), then we obtain

$$w_e^2\left(A,B\right)\ge max\left\{w\left[{\left(\alpha A+\beta B\right)}^2\right],w\left[{\left(\alpha A-\beta B\right)}^2\right]\right\},$$

(11)

for $A,B\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2=1.$ For $\alpha =\beta ={\displaystyle \frac{\sqrt{2}}{2}}$, we obtain

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

(12)

which is a known result, see [22].

If we take $\gamma =\beta ,$$\delta =\alpha$ in (5), then we obtain

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

(13)

for $A,B\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2=1.$ For $\alpha =\beta ={\displaystyle \frac{\sqrt{2}}{2}}$, we obtain again (12).

If we take $\gamma =i\alpha ,$$\delta =-i\beta$ in (5), then we obtain

$$\begin{array}{cc}\hfill w_e^2\left(A,B\right)& \ge \left|\alpha \beta \right|w\left(AB+BA\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left[\alpha \left(1+i\right)A+\beta \left(1-i\right)B\right],w\left[\alpha \left(1-i\right)A+\beta \left(1+i\right)B\right]\right\}\hfill \\ \hfill & \times \left|w\left(\alpha A+\beta B\right)-w\left(\alpha A-\beta B\right)\right|,\hfill \end{array}$$

(14)

for $A,B\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2=1.$ For $\alpha =\beta ={\displaystyle \frac{\sqrt{2}}{2}}$, we obtain

$$\begin{array}{cc}\hfill w_e^2\left(A,B\right)& \ge {\displaystyle \frac{1}{2}}w\left(AB+BA\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}max\left\{w\left[\left(1+i\right)A+\left(1-i\right)B\right],w\left[\left(1-i\right)A+\left(1+i\right)B\right]\right\}\hfill \\ \hfill & \times \left|w\left(A+B\right)-w\left(A-B\right)\right|.\hfill \end{array}$$

If we take $\gamma =i\beta ,$$\delta =-i\alpha$ in (5), then we obtain

$$\begin{array}{cc}\hfill & w_e^2\left(A,B\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}max\left\{w\left[\left({\beta }^2-{\alpha }^2\right)\left(B^2-A^2\right)+2\alpha \beta \left(AB+BA\right)\right],\right.\hfill \\ \hfill & \left.w\left[\left({\beta }^2-{\alpha }^2\right)\left(B^2-A^2\right)-2\alpha \beta \left(AB+BA\right)\right]\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left[\left(\alpha +i\beta \right)A+\left(\beta -i\alpha \right)B\right],w\left[\left(\alpha -i\beta \right)A+\left(\beta +i\alpha \right)B\right]\right\}\hfill \\ \hfill & \times \left|w\left(\alpha A+\beta B\right)-w\left(\beta A-\alpha B\right)\right|,\hfill \end{array}$$

(15)

for $A,B\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2=1.$ For $\alpha ={\displaystyle \frac{\sqrt{2}}{2}}i$ and $\beta ={\displaystyle \frac{\sqrt{2}}{2}},$ we obtain

$$w_e^2\left(A,B\right)\ge {\displaystyle \frac{1}{2}}max\left\{w\left[{\left(B+iA\right)}^2\right],w\left[{\left(B-iA\right)}^2\right]\right\}.$$

By choosing $\alpha =sin\theta ,$$\beta =cos\theta ,$$\theta \in \mathbb{R}$, we can also state the trigonometric inequalities

$$w_e^2\left(A,B\right)\ge max\left\{w\left[{\left(sin\theta A+cos\theta B\right)}^2\right],w\left[{\left(sin\theta A-cos\theta B\right)}^2\right]\right\},$$

$$\begin{array}{cc}\hfill w_e^2\left(A,B\right)& \ge {\displaystyle \frac{1}{2}}max\left\{w\left[A^2+B^2+sin\left(2\theta \right)\left(AB+BA\right)\right],\right.\hfill \\ \hfill & \left.w\left[A^2+B^2-sin\left(2\theta \right)\left(AB+BA\right)\right]\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{\left|sin\theta +cos\theta \right|w\left(A+B\right),\left|sin\theta -cos\theta \right|w\left(A-B\right)\right\}\hfill \\ \hfill & \times \left|w\left(sin\theta A+cos\theta B\right)-w\left(cos\theta A+sin\theta B\right)\right|,\hfill \end{array}$$

$$\begin{array}{cc}\hfill w_e^2\left(A,B\right)& \ge {\displaystyle \frac{1}{2}}\left|sin\left(2\theta \right)\right|w\left(AB+BA\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left[sin\theta \left(1+i\right)A+cos\theta \left(1-i\right)B\right],\right.\hfill \\ \hfill & \left.w\left[sin\theta \left(1-i\right)A+cos\theta \left(1+i\right)B\right]\right\}\hfill \\ \hfill & \times \left|w\left(sin\theta A+cos\theta B\right)-w\left(sin\theta A-cos\theta B\right)\right|,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill w_e^2\left(A,B\right)& \ge {\displaystyle \frac{1}{2}}max\left\{w\left[cos\left(2\theta \right)\left(B^2-A^2\right)+sin\left(2\theta \right)\left(AB+BA\right)\right],\right.\hfill \\ \hfill & \left.w\left[cos\left(2\theta \right)\left(B^2-A^2\right)-sin\left(2\theta \right)\left(AB+BA\right)\right]\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left[\left(sin\theta +icos\theta \right)A+\left(cos\theta -isin\theta \right)B\right],\right.\hfill \\ \hfill & \left.w\left[\left(sin\theta -icos\theta \right)A+\left(cos\theta +isin\theta \right)B\right]\right\}\hfill \\ \hfill & \times \left|w\left(sin\theta A+cos\theta B\right)-w\left(cos\theta A-sin\theta B\right)\right|,\hfill \end{array}$$

for $A,B\in \mathcal{L}\left(H\right)$ and $\theta \in \mathbb{R}$.

Now, if we take $A=T$ and $B=T^{\ast }$ in Theorem 1 and since $w_e^2\left(T,T^{\ast }\right)=2w^2\left(T\right),$ then we obtain

$$\begin{array}{cc}\hfill w^2\left(T\right)& \ge {\displaystyle \frac{1}{4}}max\left\{w\left[\left({\alpha }^2+{\gamma }^2\right)T^2+\left({\beta }^2+{\delta }^2\right){\left(T^{\ast }\right)}^2+\left(\alpha \beta +\gamma \delta \right)\left({\left|T\right|}^2+{\left|T^{\ast }\right|}^2\right)\right],\right.\hfill \\ \hfill & \left.w\left[\left({\alpha }^2+{\gamma }^2\right)T^2+\left({\beta }^2+{\delta }^2\right){\left(T^{\ast }\right)}^2-\left(\alpha \beta +\gamma \delta \right)\left({\left|T\right|}^2+{\left|T^{\ast }\right|}^2\right)\right]\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}max\left\{w\left[\left(\alpha +\gamma \right)T+\left(\beta +\delta \right)T^{\ast }\right],w\left[\left(\alpha -\gamma \right)T+\left(\beta -\delta \right)T^{\ast }\right]\right\}\hfill \\ \hfill & \times \left|w\left(\alpha T+\beta T^{\ast }\right)-w\left(\gamma T+\delta T^{\ast }\right)\right|,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill w^2\left(T\right)& \ge {\displaystyle \frac{1}{8}}max\left\{w\left[\left[{\left(\alpha +\beta \right)}^2+{\left(\gamma +\delta \right)}^2\right]T^2+\left[{\left(\alpha -\beta \right)}^2+{\left(\gamma -\delta \right)}^2\right]{\left(T^{\ast }\right)}^2\right.\right.\hfill \\ \hfill & +\left({\alpha }^2+{\gamma }^2-{\beta }^2-{\delta }^2\right)\left({\left|T\right|}^2+{\left|T^{\ast }\right|}^2\right),\hfill \\ \hfill & w\left[\left[{\left(\alpha -\beta \right)}^2+{\left(\gamma -\delta \right)}^2\right]T^2+\left[{\left(\alpha +\beta \right)}^2+{\left(\gamma +\delta \right)}^2\right]{\left(T^{\ast }\right)}^2\right.\hfill \\ \hfill & \left.+\left({\alpha }^2+{\gamma }^2-{\beta }^2-{\delta }^2\right)\left({\left|T\right|}^2+{\left|T^{\ast }\right|}^2\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{8}}max\left\{w\left[\left(\alpha +\gamma +\beta +\delta \right)T+\left(\alpha +\gamma -\beta -\delta \right)T^{\ast }\right],\right.\hfill \\ \hfill & \left.w\left[\left(\alpha +\beta -\gamma -\delta \right)T+\left(\alpha -\beta -\gamma +\delta \right)T^{\ast }\right]\right\}\hfill \\ \hfill & \times \left|w\left(\left(\alpha +\beta \right)T+\left(\alpha -\beta \right)T^{\ast }\right)-w\left(\left(\gamma +\delta \right)T+\left(\gamma -\delta \right)T^{\ast }\right)\right|.\hfill \end{array}$$

for $\alpha ,\beta ,\gamma ,\delta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2={\left|\gamma \right|}^2+{\left|\delta \right|}^2=1.$

If we take $A=T$ and $B=T^{\ast }$ and $\gamma =\alpha ,$ $\delta =\beta$ in (5), then we obtain

$$w^2\left(T\right)\ge {\displaystyle \frac{1}{2}}max\left\{w\left[{\left(\alpha T+\beta T^{\ast }\right)}^2\right],w\left[{\left(\alpha T-\beta T^{\ast }\right)}^2\right]\right\},$$

for $T\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2=1.$ For $\alpha =\beta ={\displaystyle \frac{\sqrt{2}}{2}}$, we obtain

$$w^2\left(T\right)\ge {\displaystyle \frac{1}{4}}max\left\{{∥T+T^{\ast }∥}^2,{∥T-T^{\ast }∥}^2\right\}.$$

If we take $A=T$ and $B=T^{\ast }$ and $\gamma =\beta ,$ $\delta =\alpha$ in (5), then we obtain

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

for $T\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2=1.$

For $\alpha =0$ and $\beta =1$, we obtain

$$w^2\left(T\right)\ge {\displaystyle \frac{1}{4}}w\left[\left(T^2+{\left(T^{\ast }\right)}^2\right)\right]={\displaystyle \frac{1}{4}}∥T^2+{\left(T^{\ast }\right)}^2∥,$$

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

Further, from (14), for $A=T$ and $B=T^{\ast }$, we obtain

$$\begin{array}{cc}\hfill w^2\left(T\right)& \ge {\displaystyle \frac{1}{2}}\left|\alpha \beta \right|∥{\left|T\right|}^2+{\left|T^{\ast }\right|}^2∥\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}max\left\{w\left[\alpha \left(1+i\right)T+\beta \left(1-i\right)T^{\ast }\right],w\left[\alpha \left(1-i\right)T+\beta \left(1+i\right)T^{\ast }\right]\right\}\hfill \\ \hfill & \times \left|w\left(\alpha T+\beta T^{\ast }\right)-w\left(\alpha T-\beta T^{\ast }\right)\right|,\hfill \end{array}$$

for $T\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2=1.$ For $\alpha =\beta ={\displaystyle \frac{\sqrt{2}}{2}}$, we obtain

$$\begin{array}{cc}\hfill w^2\left(T\right)& \ge {\displaystyle \frac{1}{4}}∥{\left|T\right|}^2+{\left|T^{\ast }\right|}^2∥\hfill \\ \hfill & +{\displaystyle \frac{1}{8}}max\left\{w\left[\left(1+i\right)T+\left(1-i\right)T^{\ast }\right],w\left[\left(1-i\right)T+\left(1+i\right)T^{\ast }\right]\right\}\hfill \\ \hfill & \times \left|∥T+T^{\ast }∥-∥T-T^{\ast }∥\right|,\hfill \end{array}$$

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

This inequality is better than the first inequality of Kittaneh, (2).

Moreover, if we take $A=T$ and $B=T^{\ast }$ in (15), then we obtain

$$\begin{array}{cc}\hfill & w^2\left(T\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{4}}max\left\{w\left[\left({\beta }^2-{\alpha }^2\right)\left({\left(T^{\ast }\right)}^2-T^2\right)+2\alpha \beta \left({\left|T\right|}^2+{\left|T^{\ast }\right|}^2\right)\right],\right.\hfill \\ \hfill & \left.w\left[\left({\beta }^2-{\alpha }^2\right)\left({\left(T^{\ast }\right)}^2-T^2\right)-2\alpha \beta \left({\left|T\right|}^2+{\left|T^{\ast }\right|}^2\right)\right]\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}max\left\{w\left[\left(\alpha +i\beta \right)T+\left(\beta -i\alpha \right)T^{\ast }\right],w\left[\left(\alpha -i\beta \right)T+\left(\beta +i\alpha \right)T^{\ast }\right]\right\}\hfill \\ \hfill & \times \left|w\left(\alpha T+\beta T^{\ast }\right)-w\left(\beta T-\alpha T^{\ast }\right)\right|,\hfill \end{array}$$

(16)

for $T\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2=1.$ For $\alpha ={\displaystyle \frac{\sqrt{2}}{2}}i$ and $\beta ={\displaystyle \frac{\sqrt{2}}{2}},$ we obtain

$$\begin{array}{ccc}\hfill w^2\left(T\right)& \ge & {\displaystyle \frac{1}{4}}max\left\{w\left[{\left(T^{\ast }+iT\right)}^2\right],w\left[{\left(T^{\ast }-iT\right)}^2\right]\right\}\hfill \\ & =& {\displaystyle \frac{1}{4}}max\left\{∥{\left(T^{\ast }+iT\right)}^2∥,∥{\left(T^{\ast }-iT\right)}^2∥\right\},\hfill \end{array}$$

since ${\left(T^{\ast }+iT\right)}^2$ and ${\left(T^{\ast }-iT\right)}^2$ are normal operators.

Also, if we take $\alpha =0$ and $\beta =1$ in (16), then we obtain

$$w^2\left(T\right)\ge {\displaystyle \frac{1}{4}}w\left[\left({\left(T^{\ast }\right)}^2-T^2\right)\right]={\displaystyle \frac{1}{4}}∥{\left(T^{\ast }\right)}^2-T^2∥,$$

since ${\left(T^{\ast }\right)}^2-T^2$ is normal. This inequality was obtained before in ([22], Remark 2).

If we consider the real and imaginary part of $T\in \mathcal{L}\left(H\right),$ then the following Cartesian decomposition holds: $T=\mathfrak{R}\left(T\right)+i\mathfrak{I}\left(T\right)$ and $w_e^2\left(\mathfrak{R}\left(T\right),\mathfrak{I}\left(T\right)\right)=w^2\left(T\right).$

If we write the inequality (11) for $A=\mathfrak{R}\left(T\right)$ and $B=\mathfrak{I}\left(T\right),$ then we obtain

$$w^2\left(T\right)\ge max\left\{w\left[{\left(\alpha \mathfrak{R}\left(T\right)+\beta \mathfrak{I}\left(T\right)\right)}^2\right],w\left[{\left(\alpha \mathfrak{R}\left(T\right)-\beta \mathfrak{I}\left(T\right)\right)}^2\right]\right\},$$

for $T\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2=1.$ For $\alpha =\beta ={\displaystyle \frac{\sqrt{2}}{2}}$, we obtain

$$w^2\left(T\right)\ge {\displaystyle \frac{1}{2}}max\left\{{∥\mathfrak{R}\left(T\right)+\mathfrak{I}\left(T\right)∥}^2,{∥\mathfrak{R}\left(T\right)-\mathfrak{I}\left(T\right)∥}^2\right\}.$$

By letting $A=\mathfrak{R}\left(T\right)$ and $B=\mathfrak{I}\left(T\right)$ in (13), we also have

$$\begin{array}{cc}\hfill w^2\left(T\right)& \ge {\displaystyle \frac{1}{2}}max\left\{w\left[\left({\alpha }^2+{\beta }^2\right)\left({\mathfrak{R}}^2\left(T\right)+{\mathfrak{I}}^2\left(T\right)\right)+2\alpha \beta \left(\mathfrak{R}\left(T\right)\mathfrak{I}\left(T\right)+\mathfrak{I}\left(T\right)\mathfrak{R}\left(T\right)\right)\right],\right.\hfill \\ \hfill & \left.w\left[\left({\alpha }^2+{\beta }^2\right)\left({\mathfrak{R}}^2\left(T\right)+{\mathfrak{I}}^2\left(T\right)\right)-2\alpha \beta \left(\mathfrak{R}\left(T\right)\mathfrak{I}\left(T\right)+\mathfrak{I}\left(T\right)\mathfrak{R}\left(T\right)\right)\right]\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{\left|\alpha +\beta \right|∥\mathfrak{R}\left(T\right)+\mathfrak{I}\left(T\right)∥,\left|\alpha -\beta \right|\left(∥\mathfrak{R}\left(T\right)-\mathfrak{I}\left(T\right)∥\right)\right\}\hfill \\ \hfill & \times \left|w\left(\alpha \mathfrak{R}\left(T\right)+\beta \mathfrak{I}\left(T\right)\right)-w\left(\beta \mathfrak{R}\left(T\right)+\alpha \mathfrak{I}\left(T\right)\right)\right|,\hfill \end{array}$$

for $T\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2=1.$

If we take $\alpha =0$ and $\beta =1$ in this inequality, then we obtain

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

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

Since ${\mathfrak{R}}^2\left(T\right)+{\mathfrak{I}}^2\left(T\right)={\displaystyle \frac{{\left|T\right|}^2+{\left|T^{\ast }\right|}^2}{2}},$ hence

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

which is better than the first inequality of Kittaneh, (2).

Further, if we take $A=\mathfrak{R}\left(T\right)$ and $B=\mathfrak{I}\left(T\right)$ in (14), then we obtain

$$\begin{array}{cc}\hfill w^2\left(T\right)& \ge \left|\alpha \beta \right|∥\mathfrak{R}\left(T\right)\mathfrak{I}\left(T\right)+\mathfrak{I}\left(T\right)\mathfrak{R}\left(T\right)∥\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left[\alpha \left(1+i\right)\mathfrak{R}\left(T\right)+\beta \left(1-i\right)\mathfrak{I}\left(T\right)\right],\right.\hfill \\ \hfill & \left.w\left[\alpha \left(1-i\right)\mathfrak{R}\left(T\right)+\beta \left(1+i\right)\mathfrak{I}\left(T\right)\right]\right\}\hfill \\ \hfill & \times \left|w\left(\alpha \mathfrak{R}\left(T\right)+\beta \mathfrak{I}\left(T\right)\right)-w\left(\alpha \mathfrak{R}\left(T\right)-\beta \mathfrak{I}\left(T\right)\right)\right|,\hfill \end{array}$$

for $T\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2=1.$ For $\alpha =\beta ={\displaystyle \frac{\sqrt{2}}{2}}$, we obtain

$$\begin{array}{cc}\hfill w^2\left(T\right)& \ge {\displaystyle \frac{1}{2}}∥\mathfrak{R}\left(T\right)\mathfrak{I}\left(T\right)+\mathfrak{I}\left(T\right)\mathfrak{R}\left(T\right)∥\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}max\left\{w\left[\left(1+i\right)A+\left(1-i\right)B\right],w\left[\left(1-i\right)A+\left(1+i\right)B\right]\right\}\hfill \\ \hfill & \times \left|∥\mathfrak{R}\left(T\right)+\mathfrak{I}\left(T\right)∥-∥\mathfrak{R}\left(T\right)-\mathfrak{I}\left(T\right)∥\right|,\hfill \end{array}$$

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

We can also provide some lower bounds for the Davis–Wielandt radius.

Let $T\in \mathcal{L}\left(H\right).$ If we take $A=T$ and $B={\left|T\right|}^2$ in (11), then we obtain

$$d{w}^2\left(T\right)\ge max\left\{w\left[{\left(\alpha T+\beta {\left|T\right|}^2\right)}^2\right],w\left[{\left(\alpha T-\beta {\left|T\right|}^2\right)}^2\right]\right\},$$

for $\alpha ,\beta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2=1.$ For $\alpha =\beta ={\displaystyle \frac{\sqrt{2}}{2}}$, we obtain

$$d{w}^2\left(T\right)\ge {\displaystyle \frac{1}{2}}max\left\{w\left[{\left(T+{\left|T\right|}^2\right)}^2\right],w\left[{\left(T-{\left|T\right|}^2\right)}^2\right]\right\}.$$

From (13) for $A=T$ and $B={\left|T\right|}^2$, we obtain

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

for $T\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2=1.$

Also, from (14), we obtain for the same choice of $A,B$ that

$$\begin{array}{cc}\hfill d{w}^2\left(T\right)& \ge \left|\alpha \beta \right|w\left(T{\left|T\right|}^2+{\left|T\right|}^{2}T\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left[\alpha \left(1+i\right)T+\beta \left(1-i\right){\left|T\right|}^2\right],w\left[\alpha \left(1-i\right)T+\beta \left(1+i\right){\left|T\right|}^2\right]\right\}\hfill \\ \hfill & \times \left|w\left(\alpha T+\beta {\left|T\right|}^2\right)-w\left(\alpha T-\beta {\left|T\right|}^2\right)\right|,\hfill \end{array}$$

for $T\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2=1.$ For $\alpha =\beta ={\displaystyle \frac{\sqrt{2}}{2}}$, we obtain

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

Finally, by (15), we obtain

$$\begin{array}{cc}\hfill d{w}^2\left(T\right)& \ge {\displaystyle \frac{1}{2}}max\left\{w\left[\left({\beta }^2-{\alpha }^2\right)\left({\left|T\right|}^4-T^2\right)+2\alpha \beta \left(T{\left|T\right|}^2+{\left|T\right|}^{2}T\right)\right],\right.\hfill \\ \hfill & \left.w\left[\left({\beta }^2-{\alpha }^2\right)\left({\left|T\right|}^4-T^2\right)-2\alpha \beta \left(T{\left|T\right|}^2+{\left|T\right|}^{2}T\right)\right]\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left[\left(\alpha +i\beta \right)A+\left(\beta -i\alpha \right){\left|T\right|}^2\right],w\left[\left(\alpha -i\beta \right)A+\left(\beta +i\alpha \right){\left|T\right|}^2\right]\right\}\hfill \\ \hfill & \times \left|w\left(\alpha T+\beta {\left|T\right|}^2\right)-w\left(\beta T-\alpha {\left|T\right|}^2\right)\right|,\hfill \end{array}$$

for $T\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2=1.$ For $\alpha ={\displaystyle \frac{\sqrt{2}}{2}}i$ and $\beta ={\displaystyle \frac{\sqrt{2}}{2}},$ we obtain

$$d{w}^2\left(T\right)\ge {\displaystyle \frac{1}{2}}max\left\{w\left[{\left({\left|T\right|}^2+iT\right)}^2\right],w\left[{\left({\left|T\right|}^2-iT\right)}^2\right]\right\},$$

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

From a different view point, we can also state the following result:

Theorem 2.

For $A,B\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta ,\gamma ,\delta ,\zeta ,\eta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2={\left|\gamma \right|}^2+{\left|\delta \right|}^2={\left|\zeta \right|}^2+{\left|\eta \right|}^2=1,$ we have

$$w_e^2\left(A,B\right)\ge {\displaystyle \frac{1}{2}}{w}^2\left(\left(\alpha \zeta +\gamma \eta \right)A+\left(\beta \eta +\delta \zeta \right)B\right),$$

(17)

and

$$w_e^2\left(A,B\right)\ge {\displaystyle \frac{1}{4}}{w}^2\left(\left(\left(\alpha +\delta \right)\zeta +\left(\gamma +\beta \right)\eta \right)A+\left(\left(\alpha -\delta \right)\zeta +\left(\gamma -\beta \right)\eta \right)B\right).$$

(18)

Proof. 

For ${\left|\alpha \right|}^2+{\left|\beta \right|}^2=1,$ we have

$${\left|⟨Ax,x⟩\right|}^2+{\left|⟨Bx,x⟩\right|}^2\ge {\left|⟨\left(\alpha A+\beta B\right)x,x⟩\right|}^2,$$

and for ${\left|\gamma \right|}^2+{\left|\delta \right|}^2=1,$ we have

$${\left|⟨Ax,x⟩\right|}^2+{\left|⟨Bx,x⟩\right|}^2\ge {\left|⟨\left(\gamma A+\delta B\right)x,x⟩\right|}^2.$$

If $\zeta$ and $\eta$ with ${\left|\zeta \right|}^2+{\left|\eta \right|}^2=1$, then by replacing A with $\zeta A$ and B with $\eta B$, we obtain from the first inequality above that

$${\left|\zeta \right|}^2{\left|⟨Ax,x⟩\right|}^2+{\left|\eta \right|}^2{\left|⟨Bx,x⟩\right|}^2\ge {\left|⟨\left(\alpha \zeta A+\beta \eta B\right)x,x⟩\right|}^2,$$

while by replacing A with $\eta A$ and B with $\zeta B$ in the second inequality above,

$${\left|\eta \right|}^2{\left|⟨Ax,x⟩\right|}^2+{\left|\zeta \right|}^2{\left|⟨Bx,x⟩\right|}^2\ge {\left|⟨\left(\gamma \eta A+\delta \zeta B\right)x,x⟩\right|}^2.$$

If we add these two inequalities, we obtain

$$\begin{array}{cc}\hfill {\left|⟨Ax,x⟩\right|}^2+{\left|⟨Bx,x⟩\right|}^2& \ge {\left|⟨\left(\alpha \zeta A+\beta \eta B\right)x,x⟩\right|}^2+{\left|⟨\left(\gamma \eta A+\delta \zeta B\right)x,x⟩\right|}^2,\hfill \end{array}$$

(19)

for $x\in H.$

By using the Cauchy–Schwarz inequality, we have

$$\begin{array}{cc}\hfill & {\left|⟨\left(\alpha \zeta A+\beta \eta B\right)x,x⟩\right|}^2+{\left|⟨\left(\gamma \eta A+\delta \zeta B\right)x,x⟩\right|}^2\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}{\left(\left|⟨\left(\alpha \zeta A+\beta \eta B\right)x,x⟩\right|+\left|⟨\left(\gamma \eta A+\delta \zeta B\right)x,x⟩\right|\right)}^2\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}{\left|⟨\left(\alpha \zeta A+\beta \eta B+\gamma \eta A+\delta \zeta B\right)x,x⟩\right|}^2\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}{\left|⟨\left(\left(\alpha \zeta +\gamma \eta \right)A+\left(\beta \eta +\delta \zeta \right)B\right)x,x⟩\right|}^2,\hfill \end{array}$$

(20)

for $x\in H.$

By (19) and (20), we obtain

$${\left|⟨Ax,x⟩\right|}^2+{\left|⟨Bx,x⟩\right|}^2\ge {\displaystyle \frac{1}{2}}{\left|⟨\left(\left(\alpha \zeta +\gamma \eta \right)A+\left(\beta \eta +\delta \zeta \right)B\right)x,x⟩\right|}^2,$$

for $x\in H.$

By taking the supremum in this inequality, we deduce the desired result (17).

If we write (17) for $A+B$ instead of A and $A-B$ instead of $B,$ then we obtain

$$\begin{array}{cc}\hfill w_e^2\left(A+B,A-B\right)& \ge {\displaystyle \frac{1}{2}}{w}^2\left(\left(\alpha \zeta +\gamma \eta \right)\left(A+B\right)+\left(\beta \eta +\delta \zeta \right)\left(A-B\right)\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}{w}^2\left(\left(\alpha \zeta +\gamma \eta +\beta \eta +\delta \zeta \right)A+\left(\alpha \zeta +\gamma \eta -\beta \eta -\delta \zeta \right)B\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}{w}^2\left(\left(\left(\alpha +\delta \right)\zeta +\left(\gamma +\beta \right)\eta \right)A+\left(\left(\alpha -\delta \right)\zeta +\left(\gamma -\beta \right)\eta \right)B\right),\hfill \end{array}$$

and since

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

we obtain the desired result (18). □

Remark 3.

For $\gamma =\alpha ,$$\delta =\beta$ in (17), we obtain

$$w_e^2\left(A,B\right)\ge {\displaystyle \frac{1}{2}}\left|\eta +\zeta \right|w^2\left(\alpha A+\beta B\right),$$

for $A,B\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta ,\zeta ,\eta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2={\left|\zeta \right|}^2+{\left|\eta \right|}^2=1.$

Also, by taking $\zeta =\alpha$ and $\eta =\beta$ in (17), we obtain

$$w_e^2\left(A,B\right)\ge {\displaystyle \frac{1}{2}}{w}^2\left(\left({\alpha }^2+\gamma \eta \right)A+\left({\beta }^2+\delta \zeta \right)B\right),$$

for ${\left|\alpha \right|}^2+{\left|\beta \right|}^2={\left|\gamma \right|}^2+{\left|\delta \right|}^2=1.$

By taking $\zeta =\gamma$ and $\eta =\delta$, we also have

$$w_e^2\left(A,B\right)\ge {\displaystyle \frac{1}{2}}{w}^2\left(\gamma \left(\alpha +\delta \right)A+\delta \left(\beta +\gamma \right)B\right),$$

for ${\left|\alpha \right|}^2+{\left|\beta \right|}^2={\left|\gamma \right|}^2+{\left|\delta \right|}^2=1.$

To derive a significant consequence from Theorem 2, we first recall the following lemma, as found in [23]:

Lemma 1.

Let $C,D\in \mathcal{L}\left(H\right).$ Then,

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

Now, we can state the following result:

Corollary 1.

For $A,B\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta ,\gamma ,\delta ,\zeta ,\eta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2={\left|\gamma \right|}^2+{\left|\delta \right|}^2={\left|\zeta \right|}^2+{\left|\eta \right|}^2=1,$ then

$$w_e\left(A,B\right)\ge {\displaystyle \frac{\sqrt{2}}{2}}w\left(\left[\begin{array}{cc}0& \left(\alpha \zeta +\gamma \eta \right)A\\ \left(\overline{\beta \eta }+\overline{\delta \zeta }\right)B^{\ast }& 0\end{array}\right]\right),$$

(21)

and

$$\begin{array}{cc}\hfill w_e\left(A,B\right)& \ge {\displaystyle \frac{\sqrt{2}}{4}}w\left(\left[\begin{array}{cc}0& \left(\left(\alpha +\delta \right)\zeta +\left(\gamma +\beta \right)\eta \right)A\\ \left(\left(\left(\overline{\alpha }-\overline{\delta }\right)\overline{\zeta }+\left(\overline{\gamma }-\overline{\beta }\right)\overline{\eta }\right)\right)B^{\ast }& 0\end{array}\right]\right).\hfill \end{array}$$

(22)

Proof. 

If we write the inequality (17) for $e^{i\theta }A$ instead of A and $e^{-i\theta }B$ instead of $B,$ then we obtain for $\theta \in \mathbb{R}$

$$\begin{array}{cc}\hfill w_e^2\left(A,B\right)& =w_e^2\left(e^{i\theta }A,e^{-i\theta }B\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}{w}^2\left(\left(\alpha \zeta +\gamma \eta \right)e^{i\theta }A+\left(\beta \eta +\delta \zeta \right)e^{-i\theta }B\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{8}}{∥\left(\alpha \zeta +\gamma \eta \right)e^{i\theta }A+\left(\beta \eta +\delta \zeta \right)e^{-i\theta }B∥}^2,\hfill \end{array}$$

for $A,B\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta ,\gamma ,\delta ,\zeta ,\eta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2={\left|\gamma \right|}^2+{\left|\delta \right|}^2={\left|\zeta \right|}^2+{\left|\eta \right|}^2=1.$

If we take the supremum over $\theta \in \mathbb{R}$, we obtain by Lemma 1 that

$$\begin{array}{cc}\hfill w_e^2\left(A,B\right)& \ge {\displaystyle \frac{1}{8}}\underset{\theta \in R}{sup}{∥e^{i\theta }\left(\alpha \zeta +\gamma \eta \right)A+e^{-i\theta }\left(\beta \eta +\delta \zeta \right)B∥}^2\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}{w}^2\left(\left[\begin{array}{cc}0& \left(\alpha \zeta +\gamma \eta \right)A\\ \left(\overline{\beta }\overline{\eta }+\overline{\delta }\overline{\zeta }\right)B^{\ast }& 0\end{array}\right]\right),\hfill \end{array}$$

which proves (21).

The inequality (22) follows in the same way from (18). □

Remark 4.

For $\gamma =\alpha ,$$\delta =\beta$ in (21), we obtain

$$w_e\left(A,B\right)\ge {\displaystyle \frac{\sqrt{2}}{2}}w\left(\left[\begin{array}{cc}0& \alpha \left(\zeta +\eta \right)A\\ \overline{\beta }\left(\overline{\eta }+\overline{\zeta }\right)B^{\ast }& 0\end{array}\right]\right),$$

(23)

for $A,B\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta ,\zeta ,\eta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2={\left|\zeta \right|}^2+{\left|\eta \right|}^2=1.$

For $\zeta ,\eta \in \mathbb{R}$ with ${\zeta }^2+{\eta }^2=1$, we obtain from (23) that

$$w_e\left(A,B\right)\ge {\displaystyle \frac{\sqrt{2}}{2}}\left|\zeta +\eta \right|w\left(\left[\begin{array}{cc}0& \alpha A\\ \overline{\beta }{B}^{\ast }& 0\end{array}\right]\right),$$

(24)

for $\alpha ,\beta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2=1.$

Since

$$\underset{{\zeta }^2+{\eta }^2=1}{sup}\left|\zeta +\eta \right|=\sqrt{2},$$

then from (24), we derive

$$w_e\left(A,B\right)\ge w\left(\left[\begin{array}{cc}0& \alpha A\\ \sigma B^{\ast }& 0\end{array}\right]\right),$$

(25)

for $\alpha ,\sigma \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\sigma \right|}^2=1.$

In the case of $\alpha ,\sigma \ge 0$ with ${\alpha }^2+{\sigma }^2=1$, the inequality (25) was obtained in Corollary 2.3 of [24].

Now, if we take $\sigma =\beta ={\displaystyle \frac{\sqrt{2}}{2}}$ in (25), then we obtain

$$w_e\left(A,B\right)\ge {\displaystyle \frac{\sqrt{2}}{2}}w\left(\left[\begin{array}{cc}0& A\\ B^{\ast }& 0\end{array}\right]\right).$$

If we take $A=T$ and $B=T^{\ast }$ in Theorem 2, then we obtain

$$w^2\left(T\right)\ge {\displaystyle \frac{1}{4}}{w}^2\left(\left(\alpha \zeta +\gamma \eta \right)T+\left(\beta \eta +\delta \zeta \right)T^{\ast }\right),$$

and

$$w^2\left(T\right)\ge {\displaystyle \frac{1}{8}}{w}^2\left(\left(\left(\alpha +\delta \right)\zeta +\left(\gamma +\beta \right)\eta \right)T+\left(\left(\alpha -\delta \right)\zeta +\left(\gamma -\beta \right)\eta \right)T^{\ast }\right),$$

for $T\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta ,\gamma ,\delta ,\zeta ,\eta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2={\left|\gamma \right|}^2+{\left|\delta \right|}^2={\left|\zeta \right|}^2+{\left|\eta \right|}^2=1.$

Further, if we take $A=\mathfrak{R}\left(T\right)$ and $B=\mathfrak{I}\left(T\right)$ in Theorem 2, then we obtain

$$w^2\left(T\right)\ge {\displaystyle \frac{1}{2}}{w}^2\left(\left(\alpha \zeta +\gamma \eta \right)\mathfrak{R}\left(T\right)+\left(\beta \eta +\delta \zeta \right)\mathfrak{I}\left(T\right)\right),$$

and

$$w^2\left(T\right)\ge {\displaystyle \frac{1}{4}}{w}^2\left(\left(\left(\alpha +\delta \right)\zeta +\left(\gamma +\beta \right)\eta \right)\mathfrak{R}\left(T\right)+\left(\left(\alpha -\delta \right)\zeta +\left(\gamma -\beta \right)\eta \right)\mathfrak{I}\left(T\right)\right),$$

for $T\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta ,\gamma ,\delta ,\zeta ,\eta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2={\left|\gamma \right|}^2+{\left|\delta \right|}^2={\left|\zeta \right|}^2+{\left|\eta \right|}^2=1.$

Finally, if we take $A=T$ and $B={\left|T\right|}^2$ in Theorem 2, then we obtain

$$d{w}^2\left(T\right)\ge {\displaystyle \frac{1}{2}}{w}^2\left(\left(\alpha \zeta +\gamma \eta \right)T+\left(\beta \eta +\delta \zeta \right){\left|T\right|}^2\right),$$

and

$$d{w}^2\left(T\right)\ge {\displaystyle \frac{1}{4}}{w}^2\left(\left(\left(\alpha +\delta \right)\zeta +\left(\gamma +\beta \right)\eta \right)T+\left(\left(\alpha -\delta \right)\zeta +\left(\gamma -\beta \right)\eta \right){\left|T\right|}^2\right),$$

for $T\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta ,\gamma ,\delta ,\zeta ,\eta \in \mathbb{C}$ with ${\left|\alpha \right|}^2+{\left|\beta \right|}^2={\left|\gamma \right|}^2+{\left|\delta \right|}^2={\left|\zeta \right|}^2+{\left|\eta \right|}^2=1.$

In order to derive our next main theorem, we need to recall the following generalization of Kato’s Schwarz inequality for operators obtained by F. Kittaneh in 1988 (see [25]).

Lemma 2.

If f and g are nonnegative continuous functions on $\left[0,\infty \right)$ satisfying the condition $f\left(t\right)g\left(t\right)=t$ for all $t\in \left[0,\infty \right),$ then

$$\left|⟨Tx,y⟩\right|\le ∥f\left(\left|T\right|\right)x∥∥g\left(\left|T^{\ast }\right|\right)y∥,$$

(26)

for all $T\in \mathcal{L}\left(H\right)$ and $x,y\in H.$

We observe that (26) is equivalent to

$${\left|⟨Tx,y⟩\right|}^2\le ⟨f^2\left(\left|T\right|\right)x,x⟩⟨g^2\left(\left|T^{\ast }\right|\right)y,y⟩,$$

for all $T\in \mathcal{L}\left(H\right)$ and $x,y\in H.$ Kato’s inequality is obtained for $f\left(t\right)=t^{\upsilon },$$g\left(t\right)=t^{1-\upsilon },$$\nu \in \left[0,1\right],$ i.e.,

$${\left|⟨Tx,y⟩\right|}^2\le ⟨{\left|T\right|}^{2\nu }x,x⟩⟨{\left|T^{\ast }\right|}^{2\left(1-\upsilon \right)}y,y⟩,$$

for all $T\in \mathcal{L}\left(H\right)$ and $x,y\in H.$

We can state the following result providing upper bounds for the Euclidean numerical radius:

Theorem 3.

Let f, g, h and k be nonnegative continuous functions on $\left[0,\infty \right)$ satisfying the condition $f\left(t\right)g\left(t\right)=h\left(t\right)k\left(t\right)=t$ for all $t\in \left[0,\infty \right).$ If $A,B\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta ,\gamma ,\delta \in \left\{-1,1\right\},$ then

$$w_e^2\left(A,B\right)\le w\left[\alpha f^2\left(\left|A\right|\right)+i\beta h^2\left(\left|B\right|\right)\right]w\left[\gamma g^2\left(\left|A^{\ast }\right|\right)+i\delta k^2\left(\left|B^{\ast }\right|\right)\right],$$

(27)

and

$$w_e^2\left(A,B\right)\le w\left[\alpha f^2\left(\left|A\right|\right)+i\delta k^2\left(\left|B^{\ast }\right|\right)\right]w\left[\gamma g^2\left(\left|A^{\ast }\right|\right)+i\beta h^2\left(\left|B\right|\right)\right].$$

(28)

Proof. 

We have

$$\begin{array}{cc}\hfill & {\left|⟨Ax,x⟩\right|}^2+{\left|⟨Bx,x⟩\right|}^2\hfill \\ \hfill & \le ⟨f^2\left(\left|A\right|\right)x,x⟩⟨g^2\left(\left|A^{\ast }\right|\right)x,x⟩+⟨h^2\left(\left|B\right|\right)x,x⟩⟨k^2\left(\left|B^{\ast }\right|\right)x,x⟩\hfill \\ \hfill & \le {\left({⟨f^2\left(\left|A\right|\right)x,x⟩}^2+{⟨h^2\left(\left|B\right|\right)x,x⟩}^2\right)}^{1/2}\times {\left({⟨g^2\left(\left|A^{\ast }\right|\right)x,x⟩}^2+{⟨k^2\left(\left|B^{\ast }\right|\right)x,x⟩}^2\right)}^{1/2},\hfill \end{array}$$

(29)

for $x\in H.$

Observe that for $\alpha ,\beta ,\gamma ,\delta \in \left\{-1,1\right\}$, we have

$$\begin{array}{cc}\hfill {⟨f^2\left(\left|A\right|\right)x,x⟩}^2+{⟨h^2\left(\left|B\right|\right)x,x⟩}^2& ={\left|\alpha ⟨f^2\left(\left|A\right|\right)x,x⟩+\beta ⟨h^2\left(\left|B\right|\right)x,x⟩i\right|}^2\hfill \\ \hfill & ={\left|⟨\left[\alpha f^2\left(\left|A\right|\right)+i\beta h^2\left(\left|B\right|\right)\right]x,x⟩\right|}^2,\hfill \end{array}$$

and

$${⟨g^2\left(\left|A^{\ast }\right|\right)x,x⟩}^2+{⟨k^2\left(\left|B^{\ast }\right|\right)x,x⟩}^2={\left|⟨\left[\gamma g^2\left(\left|A^{\ast }\right|\right)+i\delta k^2\left(\left|B^{\ast }\right|\right)\right]x,x⟩\right|}^2,$$

for $x\in H.$

By utilizing (29), we then obtain

$$\begin{array}{cc}\hfill {\left|⟨Ax,x⟩\right|}^2+{\left|⟨Bx,x⟩\right|}^2& \le {\left|⟨\left[\alpha f^2\left(\left|A\right|\right)+i\beta h^2\left(\left|B\right|\right)\right]x,x⟩\right|}^2{\left|⟨\left[\gamma g^2\left(\left|A^{\ast }\right|\right)+i\delta k^2\left(\left|B^{\ast }\right|\right)\right]x,x⟩\right|}^2,\hfill \end{array}$$

for $x\in H.$

Therefore,

$$\begin{array}{cc}\hfill w_e^2\left(A,B\right)& =\underset{∥x∥=1}{sup}\left({\left|⟨Ax,x⟩\right|}^2+{\left|⟨Bx,x⟩\right|}^2\right)\hfill \\ \hfill & \le \underset{∥x∥=1}{sup}\left\{{\left|⟨\left[\alpha f^2\left(\left|A\right|\right)+i\beta h^2\left(\left|B\right|\right)\right]x,x⟩\right|}^2\times {\left|⟨\left[\gamma g^2\left(\left|A^{\ast }\right|\right)+i\delta k^2\left(\left|B^{\ast }\right|\right)\right]x,x⟩\right|}^2\right\}\hfill \\ \hfill & \le \underset{∥x∥=1}{sup}{\left|⟨\left[\alpha f^2\left(\left|A\right|\right)+i\beta h^2\left(\left|B\right|\right)\right]x,x⟩\right|}^2\times \underset{∥x∥=1}{sup}{\left|⟨\left[\gamma g^2\left(\left|A^{\ast }\right|\right)+i\delta k^2\left(\left|B^{\ast }\right|\right)\right]x,x⟩\right|}^2\hfill \\ \hfill & =w\left[\alpha f^2\left(\left|A\right|\right)+i\beta h^2\left(\left|B\right|\right)\right]w\left[\gamma g^2\left(\left|A^{\ast }\right|\right)+i\delta k^2\left(\left|B^{\ast }\right|\right)\right],\hfill \end{array}$$

which proves (27).

We also have

$$\begin{array}{cc}\hfill & ⟨f^2\left(\left|A\right|\right)x,x⟩⟨g^2\left(\left|A^{\ast }\right|\right)x,x⟩+⟨h^2\left(\left|B\right|\right)x,x⟩⟨k^2\left(\left|B^{\ast }\right|\right)x,x⟩\hfill \\ \hfill & \le {\left({⟨f^2\left(\left|A\right|\right)x,x⟩}^2+{⟨k^2\left(\left|B^{\ast }\right|\right)x,x⟩}^2\right)}^{1/2}\hfill \\ \hfill & \times {\left({⟨g^2\left(\left|A^{\ast }\right|\right)x,x⟩}^2+{⟨h^2\left(\left|B\right|\right)x,x⟩}^2\right)}^{1/2},\hfill \end{array}$$

for $x\in H,$ which implies in the same way (28). □

Corollary 2.

If $A,B\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta ,\gamma ,\delta \in \left\{-1,1\right\},$ then for $\upsilon ,$$\lambda \in \left[0,1\right]$, we have

$$w_e^2\left(A,B\right)\le w\left[\alpha {\left|A\right|}^{2\upsilon }+i\beta {\left|B\right|}^{2\lambda }\right]w\left[\gamma {\left|A^{\ast }\right|}^{2\left(1-\nu \right)}+i\delta {\left|B^{\ast }\right|}^{2\left(1-\lambda \right)}\right],$$

and

$$w_e^2\left(A,B\right)\le w\left[\alpha {\left|A\right|}^{2\upsilon }+i\delta {\left|B^{\ast }\right|}^{2\left(1-\lambda \right)}\right]w\left[\gamma {\left|A^{\ast }\right|}^{2\left(1-\nu \right)}+i\beta {\left|B\right|}^{2\lambda }\right].$$

The proof follows from Theorem 3 by selecting $f\left(t\right)=t^{\upsilon },$$g\left(t\right)=t^{1-\upsilon },$$h\left(t\right)=t^{\lambda }$ and $k\left(t\right)=t^{1-\lambda }$ with $\upsilon ,$$\lambda \in \left[0,1\right].$

Remark 5.

If we take $\nu =\lambda =1/2$ in Corollary 2, then we obtain

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

for $A,B\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta ,\gamma ,\delta \in \left\{-1,1\right\}.$

If we choose $B=A^{\ast }$ in Theorem 3, then we obtain

$$w^2\left(A\right)\le {\displaystyle \frac{1}{2}}w\left[\alpha f^2\left(\left|A\right|\right)+i\beta h^2\left(\left|A^{\ast }\right|\right)\right]w\left[\gamma g^2\left(\left|A^{\ast }\right|\right)+i\delta k^2\left(\left|A\right|\right)\right],$$

and

$$w^2\left(A\right)\le {\displaystyle \frac{1}{2}}w\left[\alpha f^2\left(\left|A\right|\right)+i\delta k^2\left(\left|A\right|\right)\right]w\left[\gamma g^2\left(\left|A^{\ast }\right|\right)+i\beta h^2\left(\left|A^{\ast }\right|\right)\right],$$

where f, g, h and k are nonnegative continuous functions on $\left[0,\infty \right)$ satisfying the condition $f\left(t\right)g\left(t\right)=h\left(t\right)k\left(t\right)=t$ for all $t\in \left[0,\infty \right),$ $A,B\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta ,\gamma ,\delta \in \left\{-1,1\right\}.$

In particular, we have

$$w^2\left(A\right)\le {\displaystyle \frac{1}{2}}w\left(\alpha {\left|A\right|}^{2\upsilon }+i\beta {\left|A^{\ast }\right|}^{2\lambda }\right)w\left(\gamma {\left|A^{\ast }\right|}^{2\left(1-\nu \right)}+i\delta {\left|A\right|}^{2\left(1-\lambda \right)}\right),$$

and

$$w^2\left(A\right)\le {\displaystyle \frac{1}{2}}w\left(\alpha {\left|A\right|}^{2\upsilon }+i\delta {\left|A\right|}^{2\left(1-\lambda \right)}\right)w\left(\gamma {\left|A^{\ast }\right|}^{2\left(1-\nu \right)}+i\beta {\left|A^{\ast }\right|}^{2\lambda }\right),$$

for $\upsilon ,$$\lambda \in \left[0,1\right].$ Also, we have

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

Further, if we take $A=\mathfrak{R}\left(T\right)$ and $B=\mathfrak{I}\left(T\right)$ in Theorem 3, then we obtain

$$w^2\left(T\right)\le w\left[\alpha f^2\left(\left|\mathfrak{R}\left(T\right)\right|\right)+i\beta h^2\left(\left|\mathfrak{I}\left(T\right)\right|\right)\right]w\left[\gamma g^2\left(\left|\mathfrak{R}\left(T\right)\right|\right)+i\delta k^2\left(\left|\mathfrak{I}\left(T\right)\right|\right)\right],$$

where f, g, h and k are nonnegative continuous functions on $\left[0,\infty \right)$ satisfying the condition $f\left(t\right)g\left(t\right)=h\left(t\right)k\left(t\right)=t$ for all $t\in \left[0,\infty \right),$ $T\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta ,\gamma ,\delta \in \left\{-1,1\right\}.$

For $\upsilon ,$$\lambda \in \left[0,1\right]$, we have

$$w^2\left(T\right)\le w\left(\alpha {\left|\mathfrak{R}\left(T\right)\right|}^{2\upsilon }+i\beta {\left|\mathfrak{I}\left(T\right)\right|}^{2\lambda }\right)w\left(\gamma {\left|\mathfrak{R}\left(T\right)\right|}^{2\left(1-\nu \right)}+i\delta {\left|\mathfrak{I}\left(T\right)\right|}^{2\left(1-\lambda \right)}\right),$$

and, in particular,

$$w^2\left(T\right)\le w\left(\alpha \left|\mathfrak{R}\left(T\right)\right|+i\beta \left|\mathfrak{I}\left(T\right)\right|\right)w\left(\gamma \left|\mathfrak{R}\left(T\right)\right|+i\delta \left|\mathfrak{I}\left(T\right)\right|\right),$$

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

Moreover, if we take $A=T$ and $B={\left|T\right|}^2$ in Theorem 3, then we obtain

$$d{w}^2\left(T\right)\le w\left[\alpha f^2\left(\left|T\right|\right)+i\beta h^2\left({\left|T\right|}^2\right)\right]w\left[\gamma g^2\left(\left|T^{\ast }\right|\right)+i\delta k^2\left({\left|T\right|}^2\right)\right],$$

for $T\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta ,\gamma ,\delta \in \left\{-1,1\right\}.$

For $\upsilon ,$$\lambda \in \left[0,1\right]$, we have

$$d{w}^2\left(T\right)\le w\left(\alpha {\left|T\right|}^{2\upsilon }+i\beta {\left|T\right|}^{4\lambda }\right)w\left(\gamma {\left|T^{\ast }\right|}^{2\left(1-\nu \right)}+i\delta {\left|T\right|}^{4\left(1-\lambda \right)}\right),$$

and, in particular,

$$d{w}^2\left(T\right)\le w\left(\alpha \left|T\right|+i\beta {\left|T\right|}^2\right)w\left(\gamma \left|T^{\ast }\right|+i\delta {\left|T\right|}^2\right),$$

for $T\in \mathcal{L}\left(H\right)$ and $\alpha ,\beta ,\gamma ,\delta \in \left\{-1,1\right\}.$

We have the following Hölder’s type upper bounds as well:

Theorem 4.

Let f, $g,$ h and k be nonnegative continuous functions on $\left[0,\infty \right)$ satisfying the condition $f\left(t\right)g\left(t\right)=h\left(t\right)k\left(t\right)=t$ for all $t\in \left[0,\infty \right).$ If $A,B\in \mathcal{L}\left(H\right)$ and $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1,$ then

$$w_e^2\left(A,B\right)\le {∥f^{2p}\left(\left|A\right|\right)+h^{2p}\left(\left|B\right|\right)∥}^{1/p}{∥g^{2q}\left(\left|A^{\ast }\right|\right)+k^{2q}\left(\left|B^{\ast }\right|\right)∥}^{1/q},$$

(30)

and

$$w_e^2\left(A,B\right)\le {∥f^{2p}\left(\left|A\right|\right)+k^{2p}\left(\left|B^{\ast }\right|\right)∥}^{1/p}{∥g^{2q}\left(\left|A^{\ast }\right|\right)+h^{2q}\left(\left|B\right|\right)∥}^{1/q}.$$

(31)

In particular, we have

$$w_e^2\left(A,B\right)\le {∥f^4\left(\left|A\right|\right)+h^4\left(\left|B\right|\right)∥}^{1/2}{∥g^4\left(\left|A^{\ast }\right|\right)+k^4\left(\left|B^{\ast }\right|\right)∥}^{1/2},$$

and

$$w_e^2\left(A,B\right)\le {∥f^4\left(\left|A\right|\right)+k^4\left(\left|B^{\ast }\right|\right)∥}^{1/2}{∥g^4\left(\left|A^{\ast }\right|\right)+h^4\left(\left|B\right|\right)∥}^{1/2}.$$

Proof. 

By Hölder’s inequality for $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$, we have

$$\begin{array}{cc}\hfill & ⟨f^2\left(\left|A\right|\right)x,x⟩⟨g^2\left(\left|A^{\ast }\right|\right)x,x⟩+⟨h^2\left(\left|B\right|\right)x,x⟩⟨k^2\left(\left|B^{\ast }\right|\right)x,x⟩\hfill \\ \hfill & \le {\left({⟨f^2\left(\left|A\right|\right)x,x⟩}^p+{⟨h^2\left(\left|B\right|\right)x,x⟩}^p\right)}^{1/p}\times {\left({⟨g^2\left(\left|A^{\ast }\right|\right)x,x⟩}^q+{⟨k^2\left(\left|B^{\ast }\right|\right)x,x⟩}^q\right)}^{1/q},\hfill \end{array}$$

(32)

for $x\in H.$

By using McCharthy’s inequality for a positive operator P and power $s\ge 1,$

$${⟨Px,x⟩}^s\le ⟨P^{s}x,x⟩,∥x∥=1,$$

we have

$$\begin{array}{cc}\hfill {⟨f^2\left(\left|A\right|\right)x,x⟩}^p+{⟨h^2\left(\left|B\right|\right)x,x⟩}^p& \le ⟨f^{2p}\left(\left|A\right|\right)x,x⟩+⟨h^{2p}\left(\left|B\right|\right)x,x⟩\hfill \\ \hfill & =⟨\left(f^{2p}\left(\left|A\right|\right)+h^{2p}\left(\left|B\right|\right)\right)x,x⟩,\hfill \end{array}$$

and

$${⟨g^2\left(\left|A^{\ast }\right|\right)x,x⟩}^q+{⟨k^2\left(\left|B^{\ast }\right|\right)x,x⟩}^q\le ⟨\left(g^{2q}\left(\left|A^{\ast }\right|\right)+k^{2q}\left(\left|B^{\ast }\right|\right)\right)x,x⟩,$$

for $∥x∥=1.$

By making use of (29) and (32), we derive

$$\begin{array}{cc}\hfill {\left|⟨Ax,x⟩\right|}^2+{\left|⟨Bx,x⟩\right|}^2& \le {⟨\left(f^{2p}\left(\left|A\right|\right)+h^{2p}\left(\left|B\right|\right)\right)x,x⟩}^{1/p}{⟨\left(g^{2q}\left(\left|A^{\ast }\right|\right)+k^{2q}\left(\left|B^{\ast }\right|\right)\right)x,x⟩}^{1/q},\hfill \end{array}$$

for $∥x∥=1.$

By taking the supremum over $∥x∥=1$, we obtain the desired result (30).

We also have

$$\begin{array}{cc}\hfill & ⟨f^2\left(\left|A\right|\right)x,x⟩⟨g^2\left(\left|A^{\ast }\right|\right)x,x⟩+⟨h^2\left(\left|B\right|\right)x,x⟩⟨k^2\left(\left|B^{\ast }\right|\right)x,x⟩\hfill \\ \hfill & \le {\left({⟨f^2\left(\left|A\right|\right)x,x⟩}^p+{⟨k^2\left(\left|B^{\ast }\right|\right)x,x⟩}^p\right)}^{1/p}\times {\left({⟨g^2\left(\left|A^{\ast }\right|\right)x,x⟩}^q+{⟨h^2\left(\left|B\right|\right)x,x⟩}^q\right)}^{1/q},\hfill \end{array}$$

for $x\in H,$ which implies, as above, the inequality (31). □

Corollary 3.

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

$$w_e^2\left(A,B\right)\le {∥{\left|A\right|}^{2p\upsilon }+{\left|B\right|}^{2p\left(1-\nu \right)}∥}^{1/p}{∥{\left|A^{\ast }\right|}^{2q\lambda }+{\left|B^{\ast }\right|}^{2q\left(1-\lambda \right)}∥}^{1/q},$$

and

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

In particular, we have

$$w_e^2\left(A,B\right)\le {∥{\left|A\right|}^{4\upsilon }+{\left|B\right|}^{4\left(1-\nu \right)}∥}^{1/2}{∥{\left|A^{\ast }\right|}^{4\lambda }+{\left|B^{\ast }\right|}^{4\left(1-\lambda \right)}∥}^{1/2},$$

and

$$w_e^2\left(A,B\right)\le {∥{\left|A\right|}^{4\upsilon }+{\left|B^{\ast }\right|}^{4\left(1-\lambda \right)}∥}^{1/2}{∥{\left|A^{\ast }\right|}^{4\left(1-\nu \right)}+{\left|B\right|}^{4\lambda }∥}^{1/2}.$$

The proof follows from Theorem 4 by selecting $f\left(t\right)=t^{\upsilon },$$g\left(t\right)=t^{1-\upsilon },$$h\left(t\right)=t^{\lambda }$ and $k\left(t\right)=t^{1-\lambda }$ with $\upsilon ,$$\lambda \in \left[0,1\right].$

Remark 6.

If we take $\nu =\lambda =1/2$ in Corollary 3, then we obtain

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

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

$$w_e^2\left(A,B\right)\le {∥{\left|A\right|}^2+{\left|B\right|}^2∥}^{1/2}{∥{\left|A^{\ast }\right|}^2+{\left|B^{\ast }\right|}^2∥}^{1/2},$$

for $A,B\in \mathcal{L}\left(H\right).$

If we take $B=A^{\ast }$ in Theorem 4, then we obtain for $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$ that

$$w^2\left(A\right)\le {\displaystyle \frac{1}{2}}{∥f^{2p}\left(\left|A\right|\right)+h^{2p}\left(\left|A^{\ast }\right|\right)∥}^{1/p}{∥g^{2q}\left(\left|A^{\ast }\right|\right)+k^{2q}\left(\left|A\right|\right)∥}^{1/q},$$

and

$$w^2\left(A\right)\le {\displaystyle \frac{1}{2}}{∥f^{2p}\left(\left|A\right|\right)+k^{2p}\left(\left|A\right|\right)∥}^{1/p}{∥g^{2q}\left(\left|A^{\ast }\right|\right)+h^{2q}\left(\left|A^{\ast }\right|\right)∥}^{1/q}.$$

In particular, we have

$$w^2\left(A\right)\le {\displaystyle \frac{1}{2}}{∥f^4\left(\left|A\right|\right)+h^4\left(\left|A^{\ast }\right|\right)∥}^{1/2}{∥g^4\left(\left|A^{\ast }\right|\right)+k^4\left(\left|A\right|\right)∥}^{1/2},$$

and

$$w^2\left(A\right)\le {\displaystyle \frac{1}{2}}{∥f^4\left(\left|A\right|\right)+k^4\left(\left|A\right|\right)∥}^{1/2}{∥g^4\left(\left|A^{\ast }\right|\right)+h^4\left(\left|A^{\ast }\right|\right)∥}^{1/2}.$$

For $\upsilon ,$$\lambda \in \left[0,1\right]$, we have

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

and

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

Also, for $\upsilon =\lambda =1/2$, we have

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

In particular, we have

$$w^2\left(A\right)\le {\displaystyle \frac{1}{2}}{∥{\left|A\right|}^{4\upsilon }+{\left|A^{\ast }\right|}^{4\left(1-\nu \right)}∥}^{1/2}{∥{\left|A^{\ast }\right|}^{4\lambda }+{\left|A\right|}^{4\left(1-\lambda \right)}∥}^{1/2},$$

and

$$w^2\left(A\right)\le {\displaystyle \frac{1}{2}}{∥{\left|A\right|}^{4\upsilon }+{\left|A\right|}^{4\left(1-\lambda \right)}∥}^{1/2}{∥{\left|A^{\ast }\right|}^{4\left(1-\nu \right)}+{\left|A^{\ast }\right|}^{4\lambda }∥}^{1/2}.$$

For $\upsilon =\lambda =1/2$, we obtain the known result; see the second part of inequality (2)

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

Further, if we take $A=\mathfrak{R}\left(T\right)$ and $B=\mathfrak{I}\left(T\right)$ in Theorem 4, then we obtain for $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1,$ then

$$w^2\left(T\right)\le {∥f^{2p}\left(\left|\mathfrak{R}\left(T\right)\right|\right)+h^{2p}\left(\left|\mathfrak{I}\left(T\right)\right|\right)∥}^{1/p}{∥g^{2q}\left(\left|\mathfrak{R}\left(T\right)\right|\right)+k^{2q}\left(\left|\mathfrak{I}\left(T\right)\right|\right)∥}^{1/q},$$

and

$$w^2\left(T\right)\le {∥f^{2p}\left(\left|\mathfrak{R}\left(T\right)\right|\right)+k^{2p}\left(\left|\mathfrak{I}\left(T\right)\right|\right)∥}^{1/p}{∥g^{2q}\left(\left|\mathfrak{R}\left(T\right)\right|\right)+h^{2q}\left(\left|\mathfrak{I}\left(T\right)\right|\right)∥}^{1/q}.$$

In particular, we have

$$w^2\left(T\right)\le {∥f^4\left(\left|\mathfrak{R}\left(T\right)\right|\right)+h^4\left(\left|\mathfrak{I}\left(T\right)\right|\right)∥}^{1/2}{∥g^4\left(\left|\mathfrak{R}\left(T\right)\right|\right)+k^4\left(\left|\mathfrak{I}\left(T\right)\right|\right)∥}^{1/2},$$

and

$$w^2\left(T\right)\le {∥f^4\left(\left|\mathfrak{R}\left(T\right)\right|\right)+k^4\left(\left|\mathfrak{I}\left(T\right)\right|\right)∥}^{1/2}{∥g^4\left(\left|\mathfrak{R}\left(T\right)\right|\right)+h^4\left(\left|\mathfrak{I}\left(T\right)\right|\right)∥}^{1/2}.$$

For $\upsilon ,$$\lambda \in \left[0,1\right]$, we have

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

and

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

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

For $\upsilon =\lambda =1/2$, we obtain

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

In particular, we have

$$w^2\left(T\right)\le {∥{\left|\mathfrak{R}\left(T\right)\right|}^{4\upsilon }+{\left|\mathfrak{I}\left(T\right)\right|}^{4\left(1-\nu \right)}∥}^{1/2}{∥{\left|\mathfrak{R}\left(T\right)\right|}^{4\lambda }+{\left|\mathfrak{I}\left(T\right)\right|}^{4\left(1-\lambda \right)}∥}^{1/2},$$

and

$$w^2\left(T\right)\le {∥{\left|\mathfrak{R}\left(T\right)\right|}^{4\upsilon }+{\left|\mathfrak{I}\left(T\right)\right|}^{4\left(1-\lambda \right)}∥}^{1/2}{∥{\left|\mathfrak{R}\left(T\right)\right|}^{4\left(1-\nu \right)}+{\left|\mathfrak{I}\left(T\right)\right|}^{4\lambda }∥}^{1/2}.$$

For $\upsilon =\lambda =1/2$, we obtain

$$w^2\left(T\right)\le ∥{\left|\mathfrak{R}\left(T\right)\right|}^2+{\left|\mathfrak{I}\left(T\right)\right|}^2∥={\displaystyle \frac{1}{2}}∥{\left|T\right|}^2+{\left|T^{\ast }\right|}^2∥,$$

which is the second part of inequality (2).

Moreover, if we take $A=T$ and $B={\left|T\right|}^2$ in Theorem 4, then we obtain for $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$ that

$$d{w}^2\left(T\right)\le {∥f^{2p}\left(\left|T\right|\right)+h^{2p}\left({\left|T\right|}^2\right)∥}^{1/p}{∥g^{2q}\left(\left|T^{\ast }\right|\right)+k^{2q}\left({\left|T\right|}^2\right)∥}^{1/q},$$

and

$$d{w}^2\left(T\right)\le {∥f^{2p}\left(\left|T\right|\right)+k^{2p}\left({\left|T\right|}^2\right)∥}^{1/p}{∥g^{2q}\left(\left|T^{\ast }\right|\right)+h^{2q}\left({\left|T\right|}^2\right)∥}^{1/q}.$$

In particular, we have

$$d{w}^2\left(T\right)\le {∥f^4\left(\left|T\right|\right)+h^4\left({\left|T\right|}^2\right)∥}^{1/2}{∥g^4\left(\left|T^{\ast }\right|\right)+k^4\left({\left|T\right|}^2\right)∥}^{1/2},$$

and

$$d{w}^2\left(T\right)\le {∥f^4\left(\left|T\right|\right)+k^4\left({\left|T\right|}^2\right)∥}^{1/2}{∥g^4\left(\left|T^{\ast }\right|\right)+h^4\left({\left|T\right|}^2\right)∥}^{1/2}.$$

For $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1,$ then for $\upsilon ,$$\lambda \in \left[0,1\right]$, we have

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

and

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

In particular, we have

$$d{w}^2\left(T\right)\le {∥{\left|T\right|}^{4\upsilon }+{\left|T\right|}^{8\left(1-\nu \right)}∥}^{1/2}{∥{\left|T^{\ast }\right|}^{4\lambda }+{\left|T\right|}^{8\left(1-\lambda \right)}∥}^{1/2},$$

and

$$d{w}^2\left(T\right)\le {∥{\left|T\right|}^{4\upsilon }+{\left|T\right|}^{8\left(1-\lambda \right)}∥}^{1/2}{∥{\left|T^{\ast }\right|}^{4\left(1-\nu \right)}+{\left|T\right|}^{8\lambda }∥}^{1/2}.$$

For $\upsilon =\lambda =1/2$, we obtain

$$d{w}^2\left(T\right)\le {∥{\left|T\right|}^2+{\left|T\right|}^4∥}^{1/2}{∥{\left|T^{\ast }\right|}^2+{\left|T\right|}^4∥}^{1/2}.$$

(33)

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

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

and by (33), we obtain

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

which provides a refinement of (3) for hyponormal operators.

Theorem 5.

Let f, g, h and k be nonnegative continuous functions on $\left[0,\infty \right)$ satisfying the condition $f\left(t\right)g\left(t\right)=h\left(t\right)k\left(t\right)=t$ for all $t\in \left[0,\infty \right)$ and $A,B\in \mathcal{L}\left(H\right),$ then

$$\begin{array}{cc}\hfill w_e^2\left(A,B\right)& \le {\displaystyle \frac{1}{2}}\left[∥f^2\left(\left|A\right|\right)+h^2\left(\left|B\right|\right)∥+∥f^2\left(\left|A\right|\right)-h^2\left(\left|B\right|\right)∥\right]\times ∥g^2\left(\left|A^{\ast }\right|\right)+k^2\left(\left|B^{\ast }\right|\right)∥,\hfill \end{array}$$

(34)

and

$$\begin{array}{cc}\hfill w_e^2\left(A,B\right)& \le {\displaystyle \frac{1}{2}}\left[∥f^2\left(\left|A\right|\right)+k^2\left(\left|B^{\ast }\right|\right)∥+∥f^2\left(\left|A\right|\right)-k^2\left(\left|B^{\ast }\right|\right)∥\right]\hfill \\ \hfill & \times ∥g^2\left(\left|A^{\ast }\right|\right)+h^2\left(\left|B\right|\right)∥.\hfill \end{array}$$

(35)

Proof. 

We have

$$\begin{array}{cc}\hfill & ⟨f^2\left(\left|A\right|\right)x,x⟩⟨g^2\left(\left|A^{\ast }\right|\right)x,x⟩+⟨h^2\left(\left|B\right|\right)x,x⟩⟨k^2\left(\left|B^{\ast }\right|\right)x,x⟩\hfill \\ \hfill & \le max\left\{⟨f^2\left(\left|A\right|\right)x,x⟩,⟨h^2\left(\left|B\right|\right)x,x⟩\right\}\left(⟨g^2\left(\left|A^{\ast }\right|\right)x,x⟩+⟨k^2\left(\left|B^{\ast }\right|\right)x,x⟩\right)\hfill \\ \hfill & =max\left\{⟨f^2\left(\left|A\right|\right)x,x⟩,⟨h^2\left(\left|B\right|\right)x,x⟩\right\}⟨\left[g^2\left(\left|A^{\ast }\right|\right)+k^2\left(\left|B^{\ast }\right|\right)\right]x,x⟩,\hfill \end{array}$$

for $x\in H.$

Observe that

$$\begin{array}{cc}\hfill & max\left\{⟨f^2\left(\left|A\right|\right)x,x⟩,⟨h^2\left(\left|B\right|\right)x,x⟩\right\}\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left(⟨f^2\left(\left|A\right|\right)x,x⟩+⟨h^2\left(\left|B\right|\right)x,x⟩+\left|⟨f^2\left(\left|A\right|\right)x,x⟩-⟨h^2\left(\left|B\right|\right)x,x⟩\right|\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left(⟨\left(f^2\left(\left|A\right|\right)+h^2\left(\left|B\right|\right)\right)x,x⟩+\left|⟨\left(f^2\left(\left|A\right|\right)-h^2\left(\left|B\right|\right)\right)x,x⟩\right|\right),\hfill \end{array}$$

and by (29) and (32), we derive

$$\begin{array}{cc}\hfill {\left|⟨Ax,x⟩\right|}^2+{\left|⟨Bx,x⟩\right|}^2& \le {\displaystyle \frac{1}{2}}\left(⟨\left(f^2\left(\left|A\right|\right)+h^2\left(\left|B\right|\right)\right)x,x⟩+\left|⟨\left(f^2\left(\left|A\right|\right)-h^2\left(\left|B\right|\right)\right)x,x⟩\right|\right)\hfill \\ \hfill & \times ⟨\left[g^2\left(\left|A^{\ast }\right|\right)+k^2\left(\left|B^{\ast }\right|\right)\right]x,x⟩,\hfill \end{array}$$

for $x\in H.$

By taking the supremum over $∥x∥=1$, we deduce the desired result (34).

Also, we have

$$\begin{array}{cc}\hfill & ⟨f^2\left(\left|A\right|\right)x,x⟩⟨g^2\left(\left|A^{\ast }\right|\right)x,x⟩+⟨h^2\left(\left|B\right|\right)x,x⟩⟨k^2\left(\left|B^{\ast }\right|\right)x,x⟩\hfill \\ \hfill & \le max\left\{⟨f^2\left(\left|A\right|\right)x,x⟩,⟨k^2\left(\left|B^{\ast }\right|\right)x,x⟩\right\}\left(⟨g^2\left(\left|A^{\ast }\right|\right)x,x⟩+⟨h^2\left(\left|B\right|\right)x,x⟩\right),\hfill \end{array}$$

from which we obtain (35). □

If we take $f\left(t\right)=t^{\upsilon },$$g\left(t\right)=t^{1-\upsilon },$$h\left(t\right)=t^{\lambda }$ and $k\left(t\right)=t^{1-\lambda }$ with $\upsilon ,$$\lambda \in \left[0,1\right],$ then we obtain

Corollary 4.

If $A,B\in \mathcal{L}\left(H\right)$ and $\upsilon ,$$\lambda \in \left[0,1\right]$, then

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

and

$$\begin{array}{cc}\hfill w_e^2\left(A,B\right)& \le {\displaystyle \frac{1}{2}}\left[∥{\left|A\right|}^{2\upsilon }+{\left|B^{\ast }\right|}^{2\left(1-\lambda \right)}∥+∥{\left|A\right|}^{2\upsilon }-{\left|B^{\ast }\right|}^{2\left(1-\lambda \right)}∥\right]\times ∥{\left|A^{\ast }\right|}^{2\left(1-\nu \right)}+{\left|B\right|}^{2\lambda }∥.\hfill \end{array}$$

Remark 7.

If we take $\upsilon =\lambda =1/2$ in Corollary 4, then we obtain

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

for $A,B\in \mathcal{L}\left(H\right).$

If we take $B=A^{\ast }$ in Theorem 5, then we obtain

$$\begin{array}{cc}\hfill w^2\left(A\right)& \le {\displaystyle \frac{1}{4}}\left[∥f^2\left(\left|A\right|\right)+h^2\left(\left|A^{\ast }\right|\right)∥+∥f^2\left(\left|A\right|\right)-h^2\left(\left|A^{\ast }\right|\right)∥\right]\times ∥g^2\left(\left|A^{\ast }\right|\right)+k^2\left(\left|A\right|\right)∥,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill w^2\left(A\right)& \le {\displaystyle \frac{1}{4}}\left[∥f^2\left(\left|A\right|\right)+k^2\left(\left|A\right|\right)∥+∥f^2\left(\left|A\right|\right)-k^2\left(\left|A\right|\right)∥\right]\times ∥g^2\left(\left|A^{\ast }\right|\right)+h^2\left(\left|A^{\ast }\right|\right)∥,\hfill \end{array}$$

for $A\in \mathcal{L}\left(H\right),$ where f, g, h and k are nonnegative continuous functions on $\left[0,\infty \right)$ satisfying the condition $f\left(t\right)g\left(t\right)=h\left(t\right)k\left(t\right)=t$ for all $t\in \left[0,\infty \right).$

If in Corollary 4 we choose $B=A^{\ast },$ then we obtain

$$\begin{array}{cc}\hfill w^2\left(A\right)& \le {\displaystyle \frac{1}{4}}\left[∥{\left|A\right|}^{2\upsilon }+{\left|A^{\ast }\right|}^{2\lambda }∥+∥{\left|A\right|}^{2\upsilon }-{\left|A^{\ast }\right|}^{2\lambda }∥\right]\times ∥{\left|A^{\ast }\right|}^{2\left(1-\nu \right)}+{\left|A\right|}^{2\left(1-\lambda \right)}∥,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill w^2\left(A\right)& \le {\displaystyle \frac{1}{4}}\left[∥{\left|A\right|}^{2\upsilon }+{\left|A\right|}^{2\left(1-\lambda \right)}∥+∥{\left|A\right|}^{2\upsilon }-{\left|A\right|}^{2\left(1-\lambda \right)}∥\right]\times ∥{\left|A^{\ast }\right|}^{2\left(1-\nu \right)}+{\left|A^{\ast }\right|}^{2\lambda }∥,\hfill \end{array}$$

for $\upsilon ,$$\lambda \in \left[0,1\right].$

In particular, we have

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

Further, if we take $A=\mathfrak{R}\left(T\right)$ and $B=\mathfrak{I}\left(T\right)$ in Theorem 5, then we obtain

$$\begin{array}{cc}\hfill w^2\left(T\right)& \le {\displaystyle \frac{1}{2}}\left[∥f^2\left(\left|\mathfrak{R}\left(T\right)\right|\right)+h^2\left(\left|\mathfrak{I}\left(T\right)\right|\right)∥+∥f^2\left(\left|\mathfrak{R}\left(T\right)\right|\right)-h^2\left(\left|\mathfrak{I}\left(T\right)\right|\right)∥\right]\hfill \\ \hfill & \times ∥g^2\left(\left|\mathfrak{R}\left(T\right)\right|\right)+k^2\left(\left|\mathfrak{I}\left(T\right)\right|\right)∥,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill w^2\left(T\right)& \le {\displaystyle \frac{1}{2}}\left[∥f^2\left(\left|\mathfrak{R}\left(T\right)\right|\right)+k^2\left(\left|\mathfrak{I}\left(T\right)\right|\right)∥+∥f^2\left(\left|\mathfrak{R}\left(T\right)\right|\right)-k^2\left(\left|\mathfrak{I}\left(T\right)\right|\right)∥\right]\hfill \\ \hfill & \times ∥g^2\left(\left|\mathfrak{R}\left(T\right)\right|\right)+h^2\left(\left|\mathfrak{I}\left(T\right)\right|\right)∥,\hfill \end{array}$$

for $T\in \mathcal{L}\left(H\right),$ where f, g, h and k are nonnegative continuous functions on $\left[0,\infty \right)$ satisfying the condition $f\left(t\right)g\left(t\right)=h\left(t\right)k\left(t\right)=t$ for all $t\in \left[0,\infty \right).$

If in Corollary 4 we choose $A=\mathfrak{R}\left(T\right)$ and $B=\mathfrak{I}\left(T\right),$ then we obtain

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

and

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

for $\upsilon ,$$\lambda \in \left[0,1\right].$

In particular, we have

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

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

Finally, if we take $A=T$ and $B={\left|T\right|}^2$ in Theorem 5, then we obtain

$$\begin{array}{cc}\hfill d{w}^2\left(T\right)& \le {\displaystyle \frac{1}{2}}\left[∥f^2\left(\left|T\right|\right)+h^2\left({\left|T\right|}^2\right)∥+∥f^2\left(\left|T\right|\right)-h^2\left({\left|T\right|}^2\right)∥\right]\times ∥g^2\left(\left|T^{\ast }\right|\right)+k^2\left({\left|T\right|}^2\right)∥,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill d{w}^2\left(T\right)& \le {\displaystyle \frac{1}{2}}\left[∥f^2\left(\left|T\right|\right)+k^2\left({\left|T\right|}^2\right)∥+∥f^2\left(\left|T\right|\right)-k^2\left({\left|T\right|}^2\right)∥\right]\times ∥g^2\left(\left|T^{\ast }\right|\right)+h^2\left({\left|T\right|}^2\right)∥,\hfill \end{array}$$

for $T\in \mathcal{L}\left(H\right),$ where f, $g,$h and k are nonnegative continuous functions on $\left[0,\infty \right)$ satisfying the condition $f\left(t\right)g\left(t\right)=h\left(t\right)k\left(t\right)=t$ for all $t\in \left[0,\infty \right).$

For $\upsilon ,$$\lambda \in \left[0,1\right]$, then by Corollary 4 for $A=T$ and $B={\left|T\right|}^2$, we obtain

$$\begin{array}{cc}\hfill d{w}^2\left(T\right)& \le {\displaystyle \frac{1}{2}}\left[∥{\left|T\right|}^{2\upsilon }+{\left|T\right|}^{4\lambda }∥+∥{\left|T\right|}^{2\upsilon }-{\left|T\right|}^{4\lambda }∥\right]\times ∥{\left|T^{\ast }\right|}^{2\left(1-\nu \right)}+{\left|T\right|}^{4\left(1-\lambda \right)}∥\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill d{w}^2\left(T\right)& \le {\displaystyle \frac{1}{2}}\left[∥{\left|T\right|}^{2\upsilon }+{\left|T\right|}^{4\left(1-\lambda \right)}∥+∥{\left|T\right|}^{2\upsilon }-{\left|T\right|}^{4\left(1-\lambda \right)}∥\right]\times ∥{\left|T^{\ast }\right|}^{2\left(1-\nu \right)}+{\left|T\right|}^{4\lambda }∥.\hfill \end{array}$$

In particular,

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

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

3. Conclusions

In conclusion, the work presented in this paper marks a significant advancement in the study of the Euclidean numerical radius of pairs of operators in Hilbert spaces. By introducing novel weighted lower and upper bounds, we have not only expanded on existing results in the literature but also demonstrated the asymmetrical nature of these bounds, highlighting the complexity of the relationships explored.

Moreover, the discovery of new inequalities for the numerical radius and the Davis–Wielandt radius adds depth to our understanding of operator properties in this setting. These results pave the way for future research endeavors and open up exciting possibilities for further exploration in the field.

One promising direction for future research is the parametric extension of the Euclidean numerical radius, as defined by

$$w_v(C,D):=\underset{x\in H,\parallel x\parallel =1}{sup}{({\left|⟨Cx,x⟩\right|}^{{\displaystyle \frac{1}{v}}}+{\left|⟨Dx,x⟩\right|}^{{\displaystyle \frac{1}{v}}})}^v,(0<v<1),$$

This extension could lead to intriguing results, such as the inequality

$$\begin{array}{c}\hfill w_v(A,B)\ge max\left\{w(\alpha A+\beta B),w(\gamma A+\delta B)\right\},\end{array}$$

(36)

for $A,B\in \mathcal{L}\left(H\right)$ and complex numbers $\alpha ,\beta ,\gamma ,\delta$ satisfying

$${\left|\alpha \right|}^{{\displaystyle \frac{1}{1-v}}}+{\left|\beta \right|}^{{\displaystyle \frac{1}{1-v}}}={\left|\gamma \right|}^{{\displaystyle \frac{1}{1-v}}}+{\left|\delta \right|}^{{\displaystyle \frac{1}{1-v}}}=1,0<v<1.$$

Indeed, the application of the Hölder inequality further enriches the analysis, as it states

$$\sum _{i=1}^n|a_i||b_i|\le {\left(\sum _{i=1}^n{\left|a_i\right|}^{{\displaystyle \frac{1}{1-v}}}\right)}^{1-v}{\left(\sum _{i=1}^n{\left|b_i\right|}^{{\displaystyle \frac{1}{v}}}\right)}^v,(0<v<1).$$

So, by using Hölder inequality and carefully calculating expressions for all $x\in H$, one can establish that

$$\begin{array}{cc}\hfill {({\left|⟨Ax,x⟩\right|}^{{\displaystyle \frac{1}{v}}}+{\left|⟨Bx,x⟩\right|}^{{\displaystyle \frac{1}{v}}})}^v& ={({\left|\alpha \right|}^{{\displaystyle \frac{1}{1-v}}}+{\left|\beta \right|}^{{\displaystyle \frac{1}{1-v}}})}^{1-v}{({\left|⟨Ax,x⟩\right|}^{{\displaystyle \frac{1}{v}}}+{\left|⟨Bx,x⟩\right|}^{{\displaystyle \frac{1}{v}}})}^v\hfill \\ \hfill & \ge \left|\alpha \right|\times |⟨Ax,x⟩|+\left|\beta \right|\times |⟨Bx,x⟩|\hfill \\ \hfill & \ge |⟨\left(\alpha A+\beta B\right)x,x⟩|.\hfill \end{array}$$

Taking the supremum over $x\in H$ with $\parallel x\parallel =1$ demonstrates that

$$w_v(A,B)\ge w(\alpha A+\beta B).$$

Similarly, $w_v(A,B)\ge w(\gamma A+\delta B)$. Thus, we arrive at (36) as desired.

Therefore, the present work not only contributes significant findings to the study of operator inequalities but also serves as a promising starting point for future advancements and explorations in this field.