Some Bounds for the Generalized Spherical Numerical Radius of Operator Pairs with Applications

Abstract

This paper investigates a generalization of the spherical numerical radius for a pair $(B,C)$ of bounded linear operators on a complex Hilbert space H. The generalized spherical numerical radius is defined as $w_p(B,C):={\displaystyle \underset{x\in H,\parallel x\parallel =1}{\sup }}{\left({\left|⟨Bx,x⟩\right|}^p+{\left|⟨Cx,x⟩\right|}^p\right)}^{{\displaystyle \frac{1}{p}}}$, $p\ge 1$. We derive lower bounds for $w_p^2(B,C)$ involving combinations of B and C, where $p>1$. Additionally, we establish upper bounds in terms of operator norms. Applications include the cases where $(B,C)=(A,A^{*})$, with $A^{*}$ denoting the adjoint of a bounded linear operator A, and $(B,C)=\left(\mathfrak{R}\right(A),\mathfrak{I}(A\left)\right)$, representing the real and imaginary parts of A, respectively. We also explore applications to the so-called Davis–Wielandt p-radius for $p\ge 1$, which serves as a natural generalization of the classical Davis–Wielandt radius for Hilbert-space operators.

1. Introduction

Let $(H,\langle ·,·\rangle )$ be a complex Hilbert space, where the norm induced by the inner product $\langle ·,·\rangle$ is denoted by $\parallel ·\parallel$. The $C^{*}$-algebra of bounded linear operators on H is denoted by $\mathcal{L}\left(H\right)$. For any operator $A\in \mathcal{L}\left(H\right)$, its adjoint is represented as $A^{*}$, and we define $\left|A\right|={\left(A^{*}A\right)}^{\frac{1}{2}}$, which is the positive square root of $A^{*}A$. The real and imaginary parts of A are given by $\mathfrak{R}\left(A\right)={\displaystyle \frac{1}{2}}(A+A^{*})$ and $\mathfrak{I}\left(A\right)={\displaystyle \frac{1}{2i}}(A-A^{*})$, respectively.

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

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

while the numerical radius is defined as

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

It is well known that the numerical radius $w(·)$ defines a norm on $\mathcal{L}\left(H\right)$, which is equivalent to the operator norm $\parallel ·\parallel$. Specifically, the following inequalities hold

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

For further insights into inequalities involving norms and numerical radii, readers may refer to [1,2,3,4,5,6,7,8,9,10] and the references therein.

For a pair of bounded linear operators $(C,D)$ on H, the spherical operator radius is given by:

$$w_e(C,D):={\displaystyle \underset{\parallel x\parallel =1}{sup}}{\left({\left|\langle Cx,x\rangle \right|}^2+{\left|\langle Dx,x\rangle \right|}^2\right)}^{\frac{1}{2}}.$$

As noted in [11], $w_e:\mathcal{L}{\left(H\right)}^2\to [0,\infty )$ defines a norm, and the following inequalities hold:

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

(1)

for $(C,D)\in \mathcal{L}{\left(H\right)}^2$, where the constants ${\displaystyle \frac{\sqrt{2}}{4}}$ and 1 are optimal in (1). Additional results on the spherical operator radius can be found in [12,13,14,15] and the references therein.

Following [16] (see also [17]), the Davis–Wielandt radius of an operator $T\in \mathcal{L}\left(H\right)$, denoted by $dw\left(T\right)$, is defined as

$$dw\left(T\right)={\displaystyle \underset{\parallel x\parallel =1}{sup}}{\left({\left|\langle Tx,x\rangle \right|}^2+{\parallel Tx\parallel }^4\right)}^{\frac{1}{2}}.$$

(2)

It is evident that $dw\left(T\right)\ge 0$, and $dw\left(T\right)=0$ if and only if $T=0$. For any scalar $\varphi \in \mathbb{C}$, the following relations hold: $dw\left(\varphi T\right)>\left|\varphi \right|dw\left(T\right)$ when $\left|\varphi \right|>1$, $dw\left(\varphi T\right)<\left|\varphi \right|dw\left(T\right)$ when $\left|\varphi \right|<1$, and $dw\left(\varphi T\right)=\left|\varphi \right|dw\left(T\right)$ when $\left|\varphi \right|=1$.

Moreover, the triangle inequality $dw(T+S)\le dw\left(T\right)+dw\left(S\right)$ does not generally hold for all $T,S\in \mathcal{L}\left(H\right)$. However, it is valid when $\mathfrak{R}\left(T^{*}S\right)=0$, as shown in Corollary 2.2 in [18]. Additionally, it is straightforward to verify that:

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

Zamani and Shebrawi in Theorem 2.1 [19] demonstrated that:

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

Furthermore, in Theorems 2.13, 2.14, and 2.17 [19], the authors established the following bounds:

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

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

and

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

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

Recently, Bhunia et al. [18] [Theorem 2.4] established the following upper bound:

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

(3)

for $T\in \mathcal{L}\left(H\right)$. Additionally, in [20], the authors obtained inequalities for the sum of operators

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

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

It is worth noting that for $C=T$ and $D={\left|T\right|}^2$, we have

$$\begin{array}{c}\hfill w_e{(T,|T|}^2)=dw\left(T\right).\end{array}$$

(4)

Substituting $C=T$ and $D={\left|T\right|}^2$ into (1), we obtain:

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

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

In [21], for $p\ge 1$, the concept of a generalized numerical radius for a pair of operators $B,C\in \mathcal{L}\left(H\right)$, known as the p-spherical numerical radius, was introduced as follows:

$$w_p(B,C):={\displaystyle \underset{\parallel x\parallel =1}{sup}}{\left({\left|\langle Bx,x\rangle \right|}^p+{\left|\langle Cx,x\rangle \right|}^p\right)}^{\frac{1}{p}}.$$

This concept generalizes two important notions: the Euclidean numerical radius for $p=2$ and the so-called diamond numerical radius for $p=1$. Specifically, the diamond numerical radius is defined as

$$w(A,B):={\displaystyle \underset{∥x∥=1}{sup}}\left(\left|〈Ax,x〉\right|+\left|〈Bx,x〉\right|\right).$$

In other words, the p-spherical numerical radius $w_p(A,B)$ we tested in our manuscript interpolates Euclidean numerical radius and diamond numerical radius as special cases, so that we could give the generalized results, including the existing ones. The Davis–Wielandt radius $dw\left(T\right)$ can be derived from the Euclidean numerical radius by choosing $(A,B)=\left(T,{\left|T\right|}^2\right)$. This connection highlights that the spherical radius for $p\ge 1$ naturally extends the Davis–Wielandt radius to values beyond $p=2$.

Furthermore, in the section dedicated to applications for a single operator, the spherical radius yields several meaningful special cases. These cases provide upper bounds for the power p of the numerical radius $w^p\left(T\right)$, where $p\ge 1$. This demonstrates the versatility and utility of the spherical radius in unifying and extending various classical concepts in operator theory.

In 2017, Moslehian et al. [22] established several fundamental inequalities for the p-spherical numerical radius of a pair of operators $B,C\in \mathcal{L}\left(H\right)$, as follows:

$$w_p(B,C)\le w_q(B,C)\le 2^{\frac{1}{q}-\frac{1}{p}}{w}_p(B,C),\mathrm{for}p\ge q\ge 1,$$

$$2^{{\displaystyle \frac{1}{p}}-2}∥|B^{*}{|}^2+{\left|C^{*}\right|}^2∥\le w_p(B,C),\mathrm{for}p\ge 2,$$

$$\begin{array}{cc}\hfill 2^{{\displaystyle \frac{1}{p}}-1}max\left\{w(B+C),w(B-C)\right\}& \le w_p(B,C),\hfill \\ \hfill \mathrm{and}{2}^{{\displaystyle \frac{1}{p}}-1}max\left\{w\left(B\right),w\left(C\right)\right\}& \le w_p(B,C),\mathrm{for}p\ge 1,\hfill \end{array}$$

and

$$2^{{\displaystyle \frac{1}{p}}-1}{w}^{\frac{1}{2}}(B^2+C^2)\le w_p(B,C),\mathrm{for}p\ge 1.$$

The authors of [22] also applied their findings to the Cartesian decomposition of an operator, expressed as $A=\mathfrak{R}\left(A\right)+i\mathfrak{I}\left(A\right)$.

Inspired by (4), we naturally define the Davis–Wielandt p-radius for $p\ge 1$ as

$$\begin{array}{cc}\hfill d{w}_p\left(T\right)& :=w_p{(T,|T|}^2)={\displaystyle \underset{\parallel x\parallel =1}{sup}}{\left({\left|\langle Tx,x\rangle \right|}^p+{\left|\langle |T{|}^{2}x,x\rangle \right|}^p\right)}^{\frac{1}{p}}\hfill \\ \hfill & ={\displaystyle \underset{\parallel x\parallel =1}{sup}}{\left({\left|\langle Tx,x\rangle \right|}^p+{\parallel Tx\parallel }^{2p}\right)}^{\frac{1}{p}}.\hfill \end{array}$$

For $p=1$, we define

$$\widetilde{dw}\left(T\right):=d{w}_1\left(T\right)={\displaystyle \underset{\parallel x\parallel =1}{sup}}\left(\left|\langle Tx,x\rangle \right|+{\parallel Tx\parallel }^2\right).$$

For $p=2$, we recover the standard Davis–Wielandt radius (2).

Building on the results mentioned above, this paper introduces new lower and upper bounds for the generalized p-spherical numerical radius and the spherical numerical radius of a pair of operators $(B,C)$. We demonstrate that some of these bounds refine recent findings by other researchers. Additionally, we derive new inequalities for the Davis–Wielandt p-radius and the Davis–Wielandt radius as natural consequences. Applications for specific cases where $(B,C)=(A,A^{*})$ and $(B,C)=\left(\mathfrak{R}\right(A),\mathfrak{I}(A\left)\right)$ are also included.

Among other results, we establish the following lower bounds:

$$w_p^2(B,C)\ge w\left[{\lambda }^2{B}^2+\lambda \mu (BC+CB)+{\mu }^2{C}^2\right],$$

and

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

where $\lambda ,\mu \in \mathbb{C}$, $p,q>1$, ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$, and ${\left|\lambda \right|}^q+{\left|\mu \right|}^q=1$.

We also establish the following upper bounds:

$$\begin{array}{cc}\hfill w_p^p(B,C)& \le w_p^{\frac{p}{2}}{(|B|}^{2\varphi }{,|C|}^{2\varphi })w_p^{\frac{p}{2}}(|B^{*}{|}^{2(1-\varphi )},|C^{*}{|}^{2(1-\varphi )})\hfill \\ \hfill & \le {∥{|B|}^{2\varphi p}+{|C|}^{2\varphi p}∥}^{\frac{1}{2}}{∥|B^{*}{|}^{2(1-\varphi )p}+{|C^{*}|}^{2(1-\varphi )p}∥}^{\frac{1}{2}}\hfill \end{array}$$

for $p\ge 1$, and for $p\ge 2$ and $\lambda ,\mu ,\nu ,\xi \in \{-1,1\}$, we have

$$w_p^p(B,C)\le {w\left(\lambda \right|B|}^{\varphi p}+{i\mu \left|C\right|}^{\varphi p}\left)w\right(\nu |B^{*}{|}^{(1-\varphi )p}+i\xi |C^{*}{|}^{(1-\varphi )p})$$

where $\varphi \in [0,1]$.

2. Main Results

In this paper, we build upon the results obtained in [22] by providing several new lower bounds for $w_p(A,B)$ in terms of various families of parameters, as well as upper bounds that generalize some results from [1,2,12]. Moreover, we establish new inequalities involving families of parameters for the Davis–Wielandt radius, which are related to recent results obtained in [18,20].

In this section, we present our primary findings. We assume that $\left(B,C\right)\in \mathcal{L}{\left(H\right)}^2$ throughout this discussion. Our first result is stated as follows.

Theorem 1.

Let $\lambda ,\mu ,\nu ,\xi \in \mathbb{C}$ and $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$ and such that ${\left|\lambda \right|}^q+{\left|\mu \right|}^q={\left|\nu \right|}^q+{\left|\xi \right|}^q=1.$ Then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & w_p^2\left(B,C\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}w\left[\left({\lambda }^2+{\nu }^2\right)B^2+\left(\lambda \mu +\nu \xi \right)\left(BC+CB\right)+\left({\mu }^2+{\xi }^2\right)C^2\right]\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}w\left[\left(\lambda +\nu \right)B+\left(\mu +\xi \right)C\right]\left|w\left(\lambda B+\mu C\right)-w\left(\nu B+\xi C\right)\right|\hfill \end{array}\end{array}$$

(5)

and

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & w_p^2\left(B+C,B-C\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}w\left\{\left[{\left(\lambda +\mu \right)}^2+{\left(\nu +\xi \right)}^2\right]B^2+\left[{\left(\lambda -\mu \right)}^2+{\left(\nu -\xi \right)}^2\right]C^2\right.\hfill \\ \hfill & \left.+\left({\lambda }^2+{\nu }^2-{\mu }^2-{\xi }^2\right)\left(BC+CB\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}w\left[\left(\lambda +\nu +\mu +\xi \right)B+\left(\lambda +\nu -\mu -\xi \right)C\right]\hfill \\ \hfill & \times \left|w\left[\left(\lambda +\mu \right)B+\left(\lambda -\mu \right)C\right]-w\left[\left(\nu +\xi \right)B+\left(\nu -\xi \right)C\right]\right|.\hfill \end{array}\end{array}$$

(6)

Proof. 

Using Hölder’s inequality, we determine for $\lambda ,\mu \in \mathbb{C}$ and $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1,$ that

$$\begin{array}{cc}\hfill {\left({\left|〈Bx,x〉\right|}^p+{\left|〈Cx,x〉\right|}^p\right)}^{\frac{1}{p}}& ={\left({\left|\lambda \right|}^q+{\left|\mu \right|}^q\right)}^{\frac{1}{q}}{\left({\left|〈Bx,x〉\right|}^p+{\left|〈Cx,x〉\right|}^p\right)}^{\frac{1}{p}}\hfill \\ \hfill & \ge \left|\lambda 〈Bx,x〉+\mu 〈Cx,x〉\right|=\left|〈\left(\lambda B+\mu C\right)x,x〉\right|\hfill \end{array}$$

for all $x\in H.$

If we take the power 2 and the supremum over $x\in H,$ $∥x∥=1$, then we obtain

$$w_p^2\left(B,C\right)\ge w^2\left(\lambda B+\mu C\right).$$

Similarly, we deduce that

$$w_p^2\left(B,C\right)\ge w^2\left(\nu B+\xi C\right).$$

This implies that

$$\begin{array}{c}\hfill \begin{array}{cc}{w}_p^2\left(B,C\right)\hfill & \ge max\left\{w^2\left(\lambda B+\mu C\right),w^2\left(\nu B+\xi C\right)\right\}\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left[w^2\left[\left(\lambda B+\mu C\right)\right]+w^2\left[\left(\nu B+\xi C\right)\right]\right]\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}\left|w^2\left[\left(\lambda B+\mu C\right)\right]-w^2\left[\left(\nu B+\xi C\right)\right]\right|\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left[w^2\left[\left(\lambda B+\mu C\right)\right]+w^2\left[\left(\nu B+\xi C\right)\right]\right]\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}\left|w\left[\left(\lambda B+\mu C\right)\right]-w\left[\left(\nu B+\xi C\right)\right]\right|\times \left(w\left[\left(\lambda B+\mu C\right)\right]+w\left[\left(\nu B+\xi C\right)\right]\right).\hfill \end{array}\end{array}$$

(7)

Since

$$\begin{array}{cc}\hfill & w^2\left[\left(\lambda B+\mu C\right)\right]+w^2\left[\left(\nu B+\xi C\right)\right]\hfill \\ \hfill & \ge w\left[{\left(\lambda B+\mu C\right)}^2\right]+w\left[{\left(\nu B+\xi C\right)}^2\right]\hfill \\ \hfill & \ge w\left[{\left(\lambda B+\mu C\right)}^2+{\left(\nu B+\xi C\right)}^2\right]\hfill \\ \hfill & =w\left[{\lambda }^2{B}^2+\lambda \mu \left(BC+CB\right)+{\mu }^2{C}^2+{\nu }^2{B}^2+\nu \xi \left(BC+CB\right)+{\xi }^2{C}^2\right]\hfill \\ \hfill & =w\left[\left({\lambda }^2+{\nu }^2\right)B^2+\left(\lambda \mu +\nu \xi \right)\left(BC+CB\right)+\left({\mu }^2+{\xi }^2\right)C^2\right]\hfill \end{array}$$

and

$$\begin{array}{c}\hfill \begin{array}{cc}w\left[\left(\lambda B+\mu C\right)\right]+w\left[\left(\nu B+\xi C\right)\right]\hfill & \ge w\left(\left(\lambda B+\mu C\right)+\left(\nu B+\xi C\right)\right)\hfill \\ \hfill & =w\left[\left(\left(\lambda +\nu \right)B+\left(\mu +\xi \right)C\right)\right],\hfill \end{array}\end{array}$$

(8)

then by (7) and (8), we obtain the desired inequality (5).

Now, if we replace B by $B+C$ and C by $B-C$ in (5), then we obtain

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & w_p^2\left(B+C,B-C\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}w\left[\left({\lambda }^2+{\nu }^2\right){\left(B+C\right)}^2+\left({\mu }^2+{\xi }^2\right){\left(B-C\right)}^2\right.\hfill \\ \hfill & \left.+\left(\lambda \mu +\nu \xi \right)\left(\left(B+C\right)\left(B-C\right)+\left(B-C\right)\left(B+C\right)\right)\right]\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}w\left[\left(\left(\lambda +\nu \right)\left(B+C\right)+\left(\mu +\xi \right)\left(B-C\right)\right)\right]\hfill \\ \hfill & \left|w\left[\left(\lambda \left(B+C\right)+\mu \left(B-C\right)\right)\right]-w\left[\left(\nu \left(B+C\right)+\xi \left(B-C\right)\right)\right]\right|.\hfill \end{array}\end{array}$$

(9)

Observe that

$$\begin{array}{cc}\hfill & \left({\lambda }^2+{\nu }^2\right){\left(B+C\right)}^2+\left({\mu }^2+{\xi }^2\right){\left(B-C\right)}^2\hfill \\ \hfill & =\left({\lambda }^2+{\nu }^2\right)\left(B^2+BC+CB+C^2\right)+\left({\mu }^2+{\xi }^2\right)\left(B^2-BC-CB+C^2\right)\hfill \\ \hfill & =\left({\lambda }^2+{\nu }^2+{\mu }^2+{\xi }^2\right)B^2+\left({\lambda }^2+{\nu }^2+{\mu }^2+{\xi }^2\right)C^2\hfill \\ \hfill & +\left({\lambda }^2+{\nu }^2-{\mu }^2-{\xi }^2\right)\left(BC+CB\right)\hfill \end{array}$$

and

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

which gives that

$$\begin{array}{cc}\hfill & \left({\lambda }^2+{\nu }^2\right){\left(B+C\right)}^2+\left({\mu }^2+{\xi }^2\right){\left(B-C\right)}^2\hfill \\ \hfill & +\left(\lambda \mu +\nu \xi \right)\left(\left(B+C\right)\left(B-C\right)+\left(B-C\right)\left(B+C\right)\right)\hfill \\ \hfill & =\left({\lambda }^2+{\nu }^2+{\mu }^2+{\xi }^2\right)B^2+\left({\lambda }^2+{\nu }^2+{\mu }^2+{\xi }^2\right)C^2\hfill \\ \hfill & +\left({\lambda }^2+{\nu }^2-{\mu }^2-{\xi }^2\right)\left(BC+CB\right)+2\left(\lambda \mu +\nu \xi \right)\left(B^2-C^2\right)\hfill \\ \hfill & =\left({\lambda }^2+{\nu }^2+{\mu }^2+{\xi }^2+2\left(\lambda \mu +\nu \xi \right)\right)B^2+\left({\lambda }^2+{\nu }^2+{\mu }^2+{\xi }^2-2\left(\lambda \mu +\nu \xi \right)\right)C^2\hfill \\ \hfill & +\left({\lambda }^2+{\nu }^2-{\mu }^2-{\xi }^2\right)\left(BC+CB\right)\hfill \\ \hfill & =\left[{\left(\lambda +\mu \right)}^2+{\left(\nu +\xi \right)}^2\right]B^2+\left[{\left(\lambda -\mu \right)}^2+{\left(\nu -\xi \right)}^2\right]C^2+\left({\lambda }^2+{\nu }^2-{\mu }^2-{\xi }^2\right)\left(BC+CB\right)\hfill \end{array}$$

and from (9) we derive (6). □

Remark 1.

If we set, for example, $\lambda =\xi =0$ and $\mu =\nu =1$ in Theorem 1, then we obtain

$$w_p^2(B,C)\ge {\displaystyle \frac{1}{2}}w(B^2+C^2)+{\displaystyle \frac{1}{2}}w(B+C)\left|w\left(B\right)-w\left(C\right)\right|$$

and

$$w_p^2(B+C,B-C)\ge w(B^2+C^2)+{\displaystyle \frac{1}{2}}w\left(B\right)\left|w(B-C)-w(B+C)\right|.$$

Similar lower bounds can be derived for other choices of the parameters $\lambda ,\mu ,\nu ,\xi$, where two of them are 0 and the others are either $\pm 1$ or $\pm i$. These cases are left to the interested reader.

To illustrate our result in Theorem 1, we give a numerical example which may be helpful for the readers to understand our result. Take

$$B:=\left(\begin{array}{cc}1& -2\\ 3& 1\end{array}\right),C:=\left(\begin{array}{cc}2& -1\\ 3& 2\end{array}\right),x:=\left(\begin{array}{c}\sqrt{t}\hfill \\ \sqrt{1-t}\hfill \end{array}\right),\left(0\le t\le 1\right).$$

Then, we have

$$\begin{array}{cc}\hfill {\omega }_p(B,C)& ={\displaystyle \underset{0\le t\le 1}{max}}{\left[{\left(1+\sqrt{t(1-t)}\right)}^p+{\left(2+2\sqrt{t(1-t)}\right)}^p\right]}^{\frac{1}{p}}\hfill \\ \hfill & ={(1+2^p)}^{{\displaystyle \frac{1}{p}}}{\displaystyle \underset{0\le t\le 1}{max}}\left(1+\sqrt{t(1-t)}\right)={\displaystyle \frac{3}{2}}{(1+2^p)}^{\frac{1}{p}}.\hfill \end{array}$$

We also calculate

$$\begin{array}{cc}\hfill & \omega (B^2+C^2)={\displaystyle \underset{0\le t\le 1}{max}}|-4+10\sqrt{t(1-t)}|=4,\hfill \\ \hfill & \omega (B+C)={\displaystyle \underset{0\le t\le 1}{max}}(3+3\sqrt{t(1-t)})={\displaystyle \frac{9}{2}},\hfill \\ \hfill & \omega \left(B\right)={\displaystyle \underset{0\le t\le 1}{max}}(1+\sqrt{t(1-t)})={\displaystyle \frac{3}{2}},\hfill \\ \hfill & \omega \left(C\right)={\displaystyle \underset{0\le t\le 1}{max}}(2+2\sqrt{t(1-t)})=3.\hfill \end{array}$$

Thus, we obtain

$${\displaystyle \frac{1}{2}}\omega (B^2+C^2)+{\displaystyle \frac{1}{2}}\omega (B+C)|\omega \left(B\right)-\omega \left(C\right)|=2+{\displaystyle \frac{9}{4}}·{\displaystyle \frac{3}{2}}={\displaystyle \frac{43}{8}}.$$

Thus, the inequality

$${\omega }_p^2(B,C)\ge {\displaystyle \frac{1}{2}}\omega (B^2+C^2)+{\displaystyle \frac{1}{2}}\omega (B+C)|\omega \left(B\right)-\omega \left(C\right)|$$

(10)

becomes

$${\displaystyle \frac{9}{4}}{(1+2^p)}^{\frac{2}{p}}\ge \frac{43}{8},(p\ge 1).$$

The above inequality holds, since $p\to {(1+2^p)}^{\frac{2}{p}}$ is decreasing and ${\displaystyle \underset{p\to \infty }{lim}}{(1+a^p)}^{\frac{1}{p}}=a$ if $a>1$.

Next, we calculate

$$\begin{array}{cc}\hfill & {\omega }_p(B+C,B-C)={\displaystyle \underset{0\le t\le 1}{max}}{\left[{\left(3+3\sqrt{t(1-t)}\right)}^p+{\left(1+\sqrt{t(1-t)}\right)}^p\right]}^{\frac{1}{p}}\hfill \\ \hfill & ={\displaystyle \frac{3}{2}}{(1+3^p)}^{\frac{1}{p}},\hfill \\ \hfill & \omega (B-C)={\displaystyle \underset{0\le t\le 1}{max}}\left(3+3\sqrt{t(1-t)}\right)={\displaystyle \frac{9}{2}},\hfill \\ \hfill & \omega (B+C)={\displaystyle \underset{0\le t\le 1}{max}}|-1-\sqrt{t(1-t)}|={\displaystyle \frac{3}{2}}.\hfill \end{array}$$

Thus, the inequality

$${\omega }_p^2(B+C,B-C)\ge \omega (B^2+C^2)+{\displaystyle \frac{1}{2}}\omega \left(B\right)|\omega (B-C)-\omega (B+C)|$$

(11)

becomes

$$\begin{array}{c}\hfill {\displaystyle \frac{9}{4}}{(1+3^p)}^{\frac{2}{p}}\ge 4+{\displaystyle \frac{1}{2}}·{\displaystyle \frac{3}{2}}·\left|{\displaystyle \frac{9}{2}}-{\displaystyle \frac{3}{2}}\right|={\displaystyle \frac{25}{4}}.\end{array}$$

The above inequality is also true.

If we take a simpler example such as $B:=\alpha I,C:=\beta I,\left(\alpha ,\beta \in \mathbb{C}\right).$ Then, the inequality (10) becomes ${\left({\left|\alpha \right|}^p+{\left|\beta \right|}^p\right)}^{\frac{2}{p}}\ge {max\{|\alpha |}^2{,|\beta |}^2\}$ which is true. We can confirm that the equality holds when $\alpha =0$ or $\beta =0$. Also, the inequality (10) becomes

$${\left({|\alpha +\beta |}^p+{|\alpha -\beta |}^p\right)}^{\frac{2}{p}}\ge {\left|\alpha \right|}^2+{\left|\beta \right|}^2+{\displaystyle \frac{1}{2}}\left|\alpha \right|\left||\alpha +\beta |-|\alpha -\beta |\right|.$$

For simplicity, we consider the case $\alpha ,\beta \in \mathbb{R}$ with $\alpha ,\beta >0$. For the case $\alpha >\beta >0$, the above inequality is true since $p\to {(1+a^p)}^{\frac{1}{p}}$ is decreasing for $a>0$ and, using this, the inequality becomes ${(\alpha +\beta )}^2\ge {\alpha }^2+{\beta }^2+\alpha \beta$. For the case $\beta >\alpha >0$, the inequality similarly becomes ${(\alpha +\beta )}^2\ge 2{\alpha }^2+{\beta }^2$, which is true for $\beta >\alpha >0$.

The spherical case is of interest for applications.

Corollary 1.

Let $\lambda ,\mu ,\nu ,\xi \in \mathbb{C}$ with ${\left|\lambda \right|}^2+{\left|\mu \right|}^2={\left|\nu \right|}^2+{\left|\xi \right|}^2=1.$ Then,

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & w_e^2\left(B,C\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}w\left[\left({\lambda }^2+{\nu }^2\right)B^2+\left(\lambda \mu +\nu \xi \right)\left(BC+CB\right)+\left({\mu }^2+{\xi }^2\right)C^2\right]\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}w\left[\left(\left(\lambda +\nu \right)B+\left(\mu +\xi \right)C\right)\right]\left|w\left[\left(\lambda B+\mu C\right)\right]-w\left[\left(\nu B+\xi C\right)\right]\right|\hfill \end{array}\end{array}$$

(12)

and

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & w_e^2\left(B,C\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{4}}w\left\{\left[{\left(\lambda +\mu \right)}^2+{\left(\nu +\xi \right)}^2\right]B^2+\left[{\left(\lambda -\mu \right)}^2+{\left(\nu -\xi \right)}^2\right]C^2\right.\hfill \\ \hfill & \left.+\left({\lambda }^2+{\nu }^2-{\mu }^2-{\xi }^2\right)\left(BC+CB\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}w\left[\left(\lambda +\nu +\mu +\xi \right)B+\left(\lambda +\nu -\mu -\xi \right)C\right]\hfill \\ \hfill & \times \left|w\left[\left(\lambda +\mu \right)B+\left(\lambda -\mu \right)C\right]-w\left[\left(\nu +\xi \right)B+\left(\nu -\xi \right)C\right]\right|.\hfill \end{array}\end{array}$$

(13)

The inequality (12) follows by (5) for $p=2$ while (13) is obtained from (6) for $p=2$ and observing that

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

Remark 2.

Let $\lambda ,\mu \in \mathbb{C}$ and $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$ and such that ${\left|\lambda \right|}^q+{\left|\mu \right|}^q=1.$ If we take $\nu =\lambda$ and $\xi =\mu$ in Theorem 1, then we have

$$w_p^2\left(B,C\right)\ge w\left[{\lambda }^2{B}^2+\lambda \mu \left(BC+CB\right)+{\mu }^2{C}^2\right]$$

(14)

and

$$w_p^2\left(B+C,B-C\right)\ge w\left[{\left(\lambda +\mu \right)}^2{B}^2+{\left(\lambda -\mu \right)}^2{C}^2+\left({\lambda }^2-{\mu }^2\right)\left(BC+CB\right)\right].$$

If we take $\lambda =\mu ={\displaystyle \frac{1}{2^{\frac{1}{q}}}}$ in (14), then we obtain

$$w_p^2\left(B,C\right)\ge {\displaystyle \frac{1}{2^{2/q}}}w\left[{\left(B+C\right)}^2\right].$$

If ${\left|\lambda \right|}^2+{\left|\mu \right|}^2=1,$ then by Corollary 1

$$w_e^2\left(B,C\right)\ge w\left[{\lambda }^2{B}^2+\lambda \mu \left(BC+CB\right)+{\mu }^2{C}^2\right]$$

(15)

and

$$w_e^2\left(B,C\right)\ge {\displaystyle \frac{1}{2}}w\left[{\left(\lambda +\mu \right)}^2{B}^2+{\left(\lambda -\mu \right)}^2{C}^2+\left({\lambda }^2-{\mu }^2\right)\left(BC+CB\right)\right].$$

(16)

If we set $\nu =\mu$ and $\xi =\lambda$ in Theorem 1, then for ${\left|\lambda \right|}^q+{\left|\mu \right|}^q=1$, we deduce that

$$\begin{array}{c}\hfill \begin{array}{c}{w}_p^2\left(B,C\right)\hfill \\ \ge {\displaystyle \frac{1}{2}}w\left[\left({\lambda }^2+{\mu }^2\right)B^2+2\lambda \mu \left(BC+CB\right)+\left({\mu }^2+{\lambda }^2\right)C^2\right]\hfill \\ +{\displaystyle \frac{1}{2}}\left|\lambda +\mu \right|w\left(B+C\right)\left|w\left(\lambda B+\mu C\right)-w\left(\mu B+\lambda C\right)\right|\hfill \end{array}\end{array}$$

(17)

and

$$\begin{array}{cc}\hfill w_p^2\left(B+C,B-C\right)& \ge w\left[{\left(\lambda +\mu \right)}^2{B}^2+{\left(\lambda -\mu \right)}^2{C}^2\right]\hfill \\ \hfill & +\left|\lambda +\mu \right|w\left(B\right)\left|w\left[\left(\lambda +\mu \right)B+\left(\lambda -\mu \right)C\right]-w\left[\left(\lambda +\mu \right)B-\left(\lambda -\mu \right)C\right]\right|.\hfill \end{array}$$

Remark 3.

If ${\left|\lambda \right|}^2+{\left|\mu \right|}^2=1,$ then by Corollary 1, we have

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

(18)

and

$$\begin{array}{c}\hfill \begin{array}{c}{w}_e^2\left(B,C\right)\hfill \\ \ge {\displaystyle \frac{1}{2}}w\left[{\left(\lambda +\mu \right)}^2{B}^2+{\left(\lambda -\mu \right)}^2{C}^2\right]\hfill \\ +\left|\lambda +\mu \right|w\left(B\right)\left|w\left[\left(\lambda +\mu \right)B+\left(\lambda -\mu \right)C\right]-w\left[\left(\lambda +\mu \right)B-\left(\lambda -\mu \right)C\right]\right|.\hfill \end{array}\end{array}$$

(19)

Let $\lambda ,\mu \in \mathbb{C}$ and $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$ such that ${\left|\lambda \right|}^q+{\left|\mu \right|}^q=1.$ If we set $\nu =i\mu$ and $\xi =i\lambda$ in Theorem 1, then

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

and

$$\begin{array}{cc}\hfill & w_p^2\left(B+C,B-C\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}\left|{\lambda }^2-{\mu }^2\right|w\left(BC+CB\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}w\left[\left(1+i\right)\left(\lambda +\mu \right)B+\left(1-i\right)\left(\lambda -\mu \right)C\right]\hfill \\ \hfill & \times \left|w\left[\left(\lambda +\mu \right)B+\left(\lambda -\mu \right)C\right]-w\left[\left(\lambda +\mu \right)B-\left(\lambda -\mu \right)C\right]\right|.\hfill \end{array}$$

If ${\left|\lambda \right|}^2+{\left|\mu \right|}^2=1,$ then by Corollary 1, we deduce that

$$\begin{array}{c}\hfill \begin{array}{cc}{w}_e^2\left(B,C\right)\hfill & \ge {\displaystyle \frac{1}{2}}\left|{\lambda }^2-{\mu }^2\right|w\left(B^2-C^2\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}w\left[\left(\lambda +i\mu \right)B+\left(\mu +i\lambda \right)C\right]\left|w\left(\lambda B+\mu C\right)-w\left(\mu B+\lambda C\right)\right|\hfill \end{array}\end{array}$$

(20)

and

$$\begin{array}{c}\hfill \begin{array}{c}{w}_e^2\left(B,C\right)\hfill \\ \ge {\displaystyle \frac{1}{4}}\left|{\lambda }^2-{\mu }^2\right|w\left(BC+CB\right)\hfill \\ +{\displaystyle \frac{1}{4}}w\left[\left(1+i\right)\left(\lambda +\mu \right)B+\left(1-i\right)\left(\lambda -\mu \right)C\right]\hfill \\ \times \left|w\left[\left(\lambda +\mu \right)B+\left(\lambda -\mu \right)C\right]-w\left[\left(\lambda +\mu \right)B-\left(\lambda -\mu \right)C\right]\right|.\hfill \end{array}\end{array}$$

(21)

We have the following trigonometric inequalities as well:

$$w_e^2\left(B,C\right)\ge w\left[{cos}^2\alpha B^2+sin2\alpha \left({\displaystyle \frac{BC+CB}{2}}\right)+{sin}^2\alpha C^2\right]$$

(22)

and

$$\begin{array}{c}\hfill \begin{array}{c}{w}_e^2\left(B,C\right)\hfill \\ \ge {\displaystyle \frac{1}{2}}w\left[{\left(cos\alpha +sin\alpha \right)}^2{B}^2+{\left(cos\alpha +sin\alpha \right)}^2{C}^2+cos2\alpha \left(BC+CB\right)\right],\hfill \end{array}\end{array}$$

(23)

where $\alpha \in \mathbb{R}$.

If we take $\alpha ={\displaystyle \frac{\pi }{4}}$ in (22) and (23), then

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

and

$$w_e^2\left(B,C\right)\ge w\left(B^2+C^2\right).$$

These follow by (15) and (16), while from (18) and (19), we infer that

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& \ge {\displaystyle \frac{1}{2}}w\left[B^2+sin2\alpha \left(BC+CB\right)+C^2\right]\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}\left|cos\alpha +sin\alpha \right|w\left(B+C\right)\hfill \\ \hfill & \times \left|w\left(cos\alpha B+sin\alpha C\right)-w\left(sin\alpha B+cos\alpha C\right)\right|\hfill \end{array}$$

and

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

From (20) and (21), we also obtain

$$\begin{array}{c}\hfill \begin{array}{cc}{w}_e^2\left(B,C\right)\hfill & \ge {\displaystyle \frac{1}{2}}\left|cos2\alpha \right|w\left(B^2-C^2\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}w\left[\left(cos\alpha +isin\alpha \right)B+\left(sin\alpha +icos\alpha \right)C\right]\hfill \\ \hfill & \times \left|w\left(cos\alpha B+sin\alpha C\right)-w\left(sin\alpha B+cos\alpha C\right)\right|\hfill \end{array}\end{array}$$

(24)

and

$$\begin{array}{c}\hfill \begin{array}{c}{w}_e^2\left(B,C\right)\hfill \\ \ge {\displaystyle \frac{1}{4}}\left|cos2\alpha \right|w\left(BC+CB\right)\hfill \\ +{\displaystyle \frac{1}{4}}w\left[\left(1+i\right)\left(cos\alpha +sin\alpha \right)B+\left(1-i\right)\left(cos\alpha -sin\alpha \right)C\right]\hfill \\ \times \left|w\left[\left(cos\alpha +sin\alpha \right)B+\left(cos\alpha -sin\alpha \right)C\right]\right.\hfill \\ \left.-w\left[\left(cos\alpha +sin\alpha \right)B-\left(cos\alpha -sin\alpha \right)C\right]\right|.\hfill \end{array}\end{array}$$

(25)

If we take $\alpha =0$ in (24) and (25), then we obtain

$$w_e^2\left(B,C\right)\ge {\displaystyle \frac{1}{2}}w\left(B^2-C^2\right)+{\displaystyle \frac{1}{2}}w\left(B+iC\right)\left|w\left(B\right)-w\left(C\right)\right|$$

(26)

and

$$\begin{array}{c}\hfill \begin{array}{cc}{w}_e^2\left(B,C\right)\hfill & \ge {\displaystyle \frac{1}{4}}w\left(BC+CB\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}w\left[\left(1+i\right)B+\left(1-i\right)C\right]\left|w\left(B+C\right)-w\left(B-C\right)\right|.\hfill \end{array}\end{array}$$

(27)

If we use (26) for $iC$ instead of $C,$ then we also obtain

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

(28)

while from (27),

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & w_e^2\left(B,C\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{4}}w\left(BC+CB\right)+{\displaystyle \frac{\sqrt{2}}{4}}w\left[\left(B+C\right)\right]\left|w\left(B+iC\right)-w\left(B-iC\right)\right|.\hfill \end{array}\end{array}$$

(29)

Remark 4.

To address the sharpness question in our lower bounds above, consider, for instance, the inequality

$$w_e^2(B,C)\ge w(B^2+C^2),$$

which holds for all $B,C\in \mathcal{L}\left(H\right)$. If we set $C=B$, then we obtain the inequality

$$w^2\left(B\right)\ge w\left(B^2\right),$$

which is sharp. Indeed, if $B\ne 0$ is a self-adjoint operator, both sides of the inequality yield the same quantity ${\parallel B\parallel }^2\ne 0$.

Let us emphasize that we have shown that for self-adjoint operators $B=C$, equality is achieved in the bound

$$w_e^2(B,C)\ge w(B^2+C^2).$$

Thus, any numerical example such as

$$B=C=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right),$$

where $a,b,c$ are real numbers which will satisfy the equality, since this matrix is symmetric.

To establish some upper bounds for $w_p\left(·,·\right)$, we first need to recall Kato’s inequality [23], which is given by

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

(30)

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

Theorem 2.

For any $B,C\in \mathcal{L}\left(H\right)$ and $\varphi \in \left[0,1\right]$, we determine for $p\ge 1$ that

$$\begin{array}{c}\hfill \begin{array}{cc}{w}_p^p\left(B,C\right)\hfill & \le w_p^{\frac{p}{2}}\left({\left|B\right|}^{2\varphi },{\left|C\right|}^{2\varphi }\right)w_p^{\frac{p}{2}}\left({\left|B^{*}\right|}^{2\left(1-\varphi \right)},{\left|C^{*}\right|}^{2\left(1-\varphi \right)}\right)\hfill \\ \hfill & \le {∥{\left|B\right|}^{2\varphi p}+{\left|C\right|}^{2\varphi p}∥}^{\frac{1}{2}}{∥{\left|B^{*}\right|}^{2\left(1-\varphi \right)p}+{\left|C^{*}\right|}^{2\left(1-\varphi \right)p}∥}^{{\displaystyle \frac{1}{2}}}\hfill \end{array}\end{array}$$

(31)

and

$$w_p^p\left(B,C\right)\le {\displaystyle \frac{1}{2}}∥{\left|B\right|}^{2\varphi p}+{\left|C\right|}^{2\varphi p}+{\left|B^{*}\right|}^{2\left(1-\varphi \right)p}+{\left|C^{*}\right|}^{2\left(1-\varphi \right)p}∥.$$

(32)

Proof. 

We determine by (30) that

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

(33)

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

By the Cauchy–Schwarz inequality, we obtain

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

(34)

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

By McCarthy’s inequality for $p\ge 1,$

$${〈Px,x〉}^p\le 〈P^{p}x,x〉,P\ge 0,∥x∥=1.$$

We also have

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

and

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

for $x\in H$ with $∥x∥=1.$

Therefore, by the geometric mean-arithmetic mean inequality

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

(35)

for $x\in H$ with $∥x∥=1.$

By (33)–(35), we then obtain a sequence of inequalities as follows

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

(36)

for $x\in H$ with $∥x∥=1.$

By using (36), we deduce the desired results (31) and (32). □

We also have the following theorem.

Theorem 3.

For any $B,C\in \mathcal{L}\left(H\right)$, $\varphi \in \left[0,1\right]$, we determine for $p\ge 2$ and $\lambda ,\mu ,\nu ,\xi \in \left\{-1,1\right\}$ that

$$w_p^p\left(B,C\right)\le w\left(\lambda {\left|B\right|}^{\varphi p}+i\mu {\left|C\right|}^{\varphi p}\right)w\left(\nu {\left|B^{*}\right|}^{\left(1-\varphi \right)p}+i\xi {\left|C^{*}\right|}^{\left(1-\varphi \right)p}\right).$$

(37)

Proof. 

For $p\ge 2$, we also determine for $x\in H$ with $∥x∥=1$ that

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

(38)

Observe that for $\lambda ,\mu ,\nu ,\xi \in \left\{-1,1\right\}$

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

and, similarly

$${〈{\left|B^{*}\right|}^{\left(1-\varphi \right)p}x,x〉}^2+{〈{\left|C^{*}\right|}^{\left(1-\varphi \right)p}x,x〉}^2={\left|〈\left(\nu {\left|B^{*}\right|}^{\left(1-\varphi \right)p}+i\xi {\left|C^{*}\right|}^{\left(1-\varphi \right)p}\right)x,x〉\right|}^2.$$

By (38), we then obtain

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & {\left[{〈{\left|B\right|}^{2\varphi }x,x〉}^p+{〈{\left|C\right|}^{2\varphi }x,x〉}^p\right]}^{\frac{1}{2}}\times {\left[{〈{\left|B^{*}\right|}^{2\left(1-\varphi \right)}x,x〉}^p+{〈{\left|C^{*}\right|}^{2\left(1-\varphi \right)}x,x〉}^p\right]}^{\frac{1}{2}}\hfill \\ \hfill & \le \left|〈\left(\lambda {\left|B\right|}^{\varphi p}+i\mu {\left|C\right|}^{\varphi p}\right)x,x〉\right|\left|〈\left(\nu {\left|B^{*}\right|}^{\left(1-\varphi \right)p}+i\xi {\left|C^{*}\right|}^{\left(1-\varphi \right)p}\right)x,x〉\right|\hfill \end{array}\end{array}$$

(39)

for $x\in H$ with $∥x∥=1.$

From (33), (34) and (39), we obtain

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

for $x\in H$ with $∥x∥=1.$

By taking the supremum over $x\in H$ with $∥x∥=1$ in the last inequality, we obtain (37). □

Also, recall the following result for operator matrices, established by F. Kittaneh in [24].

Lemma 1.

Let $A,$ $B,$ $C\in \mathcal{L}\left(H\right)$ with $A,$ $B\ge 0.$ Then, the operator matrix

$$\left[\begin{array}{cc}A& C^{*}\\ C& B\end{array}\right]\in \mathcal{L}\left(H\oplus H\right)$$

is positive, if and only if

$${\left|〈Cx,y〉\right|}^2\le 〈Ax,x〉〈By,y〉$$

for all $x$, $y\in H.$

Using this lemma, we can also state the following result.

Theorem 4.

Let $A,$ $B,$ $C, D, E, F\in \mathcal{L}\left(H\right)$ with $A,$ $B$, $D$, $E\ge 0$ and

$$\left[\begin{array}{cc}A& C^{*}\\ C& B\end{array}\right],\left[\begin{array}{cc}D& F^{*}\\ F& E\end{array}\right]\in \mathcal{L}\left(H\oplus H\right)$$

are positive. Then, for $p\ge 1$

$$w_p^p\left(C,F\right)\le w_p^{\frac{p}{2}}\left(A,D\right)w_p^{\frac{p}{2}}\left(B,E\right)\le {∥A^p+D^p∥}^{\frac{1}{2}}{∥B^p+E^p∥}^{\frac{1}{2}}$$

(40)

and

$$w_p^p\left(C,F\right)\le {\displaystyle \frac{1}{2}}∥A^p+D^p+B^p+E^p∥.$$

(41)

Proof. 

From Lemma 1, we have

$$\left|〈Cx,x〉\right|\le {〈Ax,x〉}^{\frac{1}{2}}{〈Bx,x〉}^{\frac{1}{2}}$$

and

$$\left|〈Fx,x〉\right|\le {〈Dx,x〉}^{\frac{1}{2}}{〈Ex,x〉}^{\frac{1}{2}}$$

for all $x\in H.$

Therefore, we obtain

$$\begin{array}{cc}\hfill {\left|〈Cx,x〉\right|}^p+{\left|〈Fx,x〉\right|}^p& \le {〈Ax,x〉}^{\frac{p}{2}}{〈Bx,x〉}^{\frac{p}{2}}+{〈Dx,x〉}^{\frac{p}{2}}{〈Ex,x〉}^{\frac{p}{2}}\hfill \\ \hfill & \le {\left({〈Ax,x〉}^p+{〈Dx,x〉}^p\right)}^{\frac{1}{2}}{\left({〈Bx,x〉}^p+{〈Ex,x〉}^p\right)}^{\frac{1}{2}}\hfill \end{array}$$

for all $x\in H.$

We have

$$\begin{array}{cc}\hfill w_p^p\left(C,F\right)& ={\displaystyle \underset{∥x∥=1}{sup}}\left({\left|〈Cx,x〉\right|}^p+{\left|〈Fx,x〉\right|}^p\right)\hfill \\ \hfill & \le {\displaystyle \underset{∥x∥=1}{sup}}\left[{\left({〈Ax,x〉}^p+{〈Dx,x〉}^p\right)}^{\frac{1}{2}}{\left({〈Bx,x〉}^p+{〈Ex,x〉}^p\right)}^{\frac{1}{2}}\right]\hfill \\ \hfill & \le {\displaystyle \underset{∥x∥=1}{sup}}{\left({〈Ax,x〉}^p+{〈Dx,x〉}^p\right)}^{\frac{1}{2}}{\displaystyle \underset{∥x∥=1}{sup}}{\left({〈Bx,x〉}^p+{〈Ex,x〉}^p\right)}^{\frac{1}{2}}\hfill \\ \hfill & =w_p^{\frac{p}{2}}\left(A,D\right)w_p^{\frac{p}{2}}\left(B,E\right),\hfill \end{array}$$

which proves the first inequality in (40).

By McCarthy’s inequality, we also have for $x\in H,$ $∥x∥=1$ that

$${\left({〈Ax,x〉}^p+{〈Dx,x〉}^p\right)}^{\frac{1}{2}}\le {\left(〈A^{p}x,x〉+〈D^{p}x,x〉\right)}^{\frac{1}{2}}={〈\left(A^p+D^p\right)x,x〉}^{\frac{1}{2}}$$

and

$${\left({〈Bx,x〉}^p+{〈Ex,x〉}^p\right)}^{\frac{1}{2}}\le {〈\left(B^p+E^p\right)x,x〉}^{\frac{1}{2}}$$

giving that

$$w_p^{\frac{p}{2}}\left(A,D\right)\le {∥A^p+D^p∥}^{\frac{1}{2}}\mathrm{and}{w}_p^{\frac{p}{2}}\left(B,E\right)\le {∥B^p+E^p∥}^{\frac{1}{2}},$$

which proves the second part of (40).

We also observe from the above that for $x\in H,$ $∥x∥=1,$

$$\begin{array}{cc}\hfill {\left|〈Cx,x〉\right|}^p+{\left|〈Fx,x〉\right|}^p& \le {〈\left(A^p+D^p\right)x,x〉}^{\frac{1}{2}}{〈\left(B^p+E^p\right)x,x〉}^{\frac{1}{2}}\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left[〈\left(A^p+D^p\right)x,x〉+〈\left(B^p+E^p\right)x,x〉\right]\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}〈\left(A^p+D^p+B^p+E^p\right)x,x〉,\hfill \end{array}$$

which, by taking the supremum over $x\in H,$ $∥x∥=1$, gives (41). □

Remark 5.

Under the assumptions of Theorem 4, we observe that for $p=1$, we obtain

$$w\left(C,F\right)\le w^{\frac{1}{2}}\left(A,D\right)w^{\frac{1}{2}}\left(B,E\right)\le {∥A+D∥}^{\frac{1}{2}}{∥B+E∥}^{\frac{1}{2}}$$

and

$$w\left(C,F\right)\le {\displaystyle \frac{1}{2}}∥A+D+B+E∥.$$

For $p=2$, we obtain the following inequalities for the spherical numerical radius:

$$w_e^2\left(C,F\right)\le w_e\left(A,D\right)w_e\left(B,E\right)\le {∥A^2+D^2∥}^{\frac{1}{2}}{∥B^2+E^2∥}^{\frac{1}{2}}$$

and

$$w_e^2\left(C,F\right)\le {\displaystyle \frac{1}{2}}∥A^2+D^2+B^2+E^2∥.$$

Corollary 2.

For $S,T,V,W\in \mathcal{L}\left(H\right)$ and $p\ge 1$, we have

$$\begin{array}{cc}\hfill w_p^p\left(SV,TW\right)& \le w_p^{\frac{p}{2}}\left({\left|S^{*}\right|}^2,{\left|T^{*}\right|}^2\right)w_p^{\frac{p}{2}}\left({\left|V\right|}^2,{\left|W\right|}^2\right)\hfill \\ \hfill & \le {∥{\left|S^{*}\right|}^{2p}+{\left|T^{*}\right|}^{2p}∥}^{\frac{1}{2}}{∥{\left|V\right|}^{2p}+{\left|W\right|}^{2p}∥}^{\frac{1}{2}}\hfill \end{array}$$

and

$$w_p^p\left(SV,TW\right)\le {\displaystyle \frac{1}{2}}∥{\left|S^{*}\right|}^{2p}+{\left|T^{*}\right|}^{2p}+{\left|V\right|}^{2p}+{\left|W\right|}^{2p}∥.$$

Proof. 

Observe that the operator matrices

$$\left[\begin{array}{cc}S{S}^{*}& SV\\ V^{*}{S}^{*}& V^{*}V\end{array}\right]\in \mathcal{L}\left(H\oplus H\right)$$

and

$$\left[\begin{array}{cc}T{T}^{*}& TW\\ W^{*}{T}^{*}& W^{*}W\end{array}\right]\in \mathcal{L}\left(H\oplus H\right)$$

are positive. The proof follows now by Theorem 4 by taking $A={\left|S^{*}\right|}^2,$ $B={\left|V\right|}^2,$ $C=SV,$ $D={\left|T^{*}\right|}^2,$ $E={\left|W\right|}^2$ and $F=TW.$ □

Remark 6.

Under the assumptions of Corollary 2, we observe that for $p=1$, we obtain

$$\begin{array}{cc}\hfill w\left(SV,TW\right)& \le w^{\frac{1}{2}}\left({\left|S^{*}\right|}^2,{\left|T^{*}\right|}^2\right)w^{\frac{1}{2}}\left({\left|V\right|}^2,{\left|W\right|}^2\right)\hfill \\ \hfill & \le {∥{\left|S^{*}\right|}^2+{\left|T^{*}\right|}^2∥}^{\frac{1}{2}}{∥{\left|V\right|}^2+{\left|W\right|}^2∥}^{\frac{1}{2}}\hfill \end{array}$$

and

$$w\left(SV,TW\right)\le {\displaystyle \frac{1}{2}}∥{\left|S^{*}\right|}^2+{\left|T^{*}\right|}^2+{\left|V\right|}^2+{\left|W\right|}^2∥.$$

For $p=2$, we obtain

$$\begin{array}{cc}\hfill w_e^2\left(SV,TW\right)& \le w_e\left({\left|S^{*}\right|}^2,{\left|T^{*}\right|}^2\right)w_e\left({\left|V\right|}^2,{\left|W\right|}^2\right)\hfill \\ \hfill & \le {∥{\left|S^{*}\right|}^4+{\left|T^{*}\right|}^4∥}^{\frac{1}{2}}{∥{\left|V\right|}^4+{\left|W\right|}^4∥}^{\frac{1}{2}}\hfill \end{array}$$

and

$$w_e^2\left(SV,TW\right)\le {\displaystyle \frac{1}{2}}∥{\left|S^{*}\right|}^4+{\left|T^{*}\right|}^4+{\left|V\right|}^4+{\left|W\right|}^4∥.$$

From a different perspective, we also have the following theorem:

Theorem 5.

Let $A,$ $B,$ $C,$ $D,$ $E,$ $F\in \mathcal{L}\left(H\right)$ with $A,$ $B,$ $D,$ $E\ge 0$ and

$$\left[\begin{array}{cc}A& C^{*}\\ C& B\end{array}\right],\left[\begin{array}{cc}D& F^{*}\\ F& E\end{array}\right]\in \mathcal{L}\left(H\oplus H\right)$$

are positive. Then,

$$w_p^p\left(C,F\right)\le w\left(\lambda A^{\frac{p}{2}}+i\mu D^{\frac{p}{2}}\right)w\left(\nu B^{\frac{p}{2}}+i\xi E^{\frac{p}{2}}\right),$$

(42)

where $p\ge 2$ and $\lambda ,\mu ,\nu ,\xi \in \left\{-1,1\right\}.$

In particular, the following inequality for the spherical numerical radius holds

$$w_e^2\left(C,F\right)\le w\left(\lambda A+i\mu D\right)w\left(\nu B+i\xi E\right).$$

Proof. 

We also determine for $p\ge 2$ and $x\in H,$ $∥x∥=1$ that

$$\begin{array}{c}\hfill \begin{array}{c}{\left|〈Cx,x〉\right|}^p+{\left|〈Fx,x〉\right|}^p\hfill \\ \le {\left({〈Ax,x〉}^p+{〈Dx,x〉}^p\right)}^{\frac{1}{2}}{\left({〈Bx,x〉}^p+{〈Ex,x〉}^p\right)}^{\frac{1}{2}}\hfill \\ ={\left({〈Ax,x〉}^{2\frac{p}{2}}+{〈Dx,x〉}^{2\frac{p}{2}}\right)}^{\frac{1}{2}}{\left({〈Bx,x〉}^{2\frac{p}{2}}+{〈Ex,x〉}^{2\frac{p}{2}}\right)}^{\frac{1}{2}}\hfill \\ \le {\left({〈A^{\frac{p}{2}}x,x〉}^2+{〈D^{\frac{p}{2}}x,x〉}^2\right)}^{\frac{1}{2}}{\left({〈B^{\frac{p}{2}}x,x〉}^2+{〈E^{\frac{p}{2}}x,x〉}^2\right)}^{\frac{1}{2}}.\hfill \end{array}\end{array}$$

(43)

Observe that for $\lambda ,\mu ,\nu ,\xi \in \left\{-1,1\right\},$

$$\begin{array}{ccc}\hfill {〈A^{\frac{p}{2}}x,x〉}^2+{〈D^{\frac{p}{2}}x,x〉}^2& =& {〈\lambda A^{\frac{p}{2}}x,x〉}^2+{〈\mu D^{\frac{p}{2}}x,x〉}^2\hfill \\ & =& {\left|〈\lambda A^{\frac{p}{2}}x,x〉+i〈\mu D^{\frac{p}{2}}x,x〉\right|}^2\hfill \\ & =& {\left|〈\left(\lambda A^{\frac{p}{2}}+i\mu D^{\frac{p}{2}}\right)x,x〉\right|}^2\hfill \end{array}$$

and similarly,

$${〈B^{\frac{p}{2}}x,x〉}^2+{〈E^{\frac{p}{2}}x,x〉}^2={\left|〈\left(\nu B^{\frac{p}{2}}+i\xi E^{\frac{p}{2}}\right)x,x〉\right|}^2$$

for $x\in H$, $∥x∥=1.$

By (43), we obtain

$${\left|〈Cx,x〉\right|}^p+{\left|〈Fx,x〉\right|}^p\le \left|〈\left(\lambda A^{\frac{p}{2}}+i\mu D^{\frac{p}{2}}\right)x,x〉\right|\left|〈\left(\nu B^{\frac{p}{2}}+i\xi E^{\frac{p}{2}}\right)x,x〉\right|$$

for $x\in H$, $∥x∥=1.$

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

$$\begin{array}{cc}\hfill w_p^p\left(C,F\right)& ={\displaystyle \underset{∥x∥=1}{sup}}{\left|〈Cx,x〉\right|}^p+{\left|〈Fx,x〉\right|}^p\hfill \\ \hfill & \le {\displaystyle \underset{∥x∥=1}{sup}}\left[\left|〈\left(\lambda A^{\frac{p}{2}}+i\mu D^{\frac{p}{2}}\right)x,x〉\right|\left|〈\left(\nu B^{\frac{p}{2}}+i\xi E^{\frac{p}{2}}\right)x,x〉\right|\right]\hfill \\ \hfill & \le {\displaystyle \underset{∥x∥=1}{sup}}\left|〈\left(\lambda A^{\frac{p}{2}}+i\mu D^{\frac{p}{2}}\right)x,x〉\right|\times {\displaystyle \underset{∥x∥=1}{sup}}\left|〈\left(\nu B^{\frac{p}{2}}+i\xi E^{\frac{p}{2}}\right)x,x〉\right|\hfill \\ \hfill & =w\left(\lambda A^{\frac{p}{2}}+i\mu D^{\frac{p}{2}}\right)w\left(\nu B^{\frac{p}{2}}+i\xi E^{\frac{p}{2}}\right),\hfill \end{array}$$

which proves (42). □

We can also state the following corollary.

Corollary 3.

For $S,T,V,W\in \mathcal{L}\left(H\right)$ and $p\ge 2$, we have

$$w_p^p\left(SV,TW\right)\le w\left(\lambda {\left|S^{*}\right|}^p+i\mu {\left|T^{*}\right|}^p\right)w\left(\nu {\left|V\right|}^p+i\xi {\left|W\right|}^p\right),$$

where $p\ge 2$ and $\lambda ,\mu ,\nu ,\xi \in \left\{-1,1\right\}.$

In particular, the following inequality for the spherical numerical radius holds:

$$w_e^2\left(SV,TW\right)\le w\left(\lambda {\left|S^{*}\right|}^2+i\mu {\left|T^{*}\right|}^2\right)w\left(\nu {\left|V\right|}^2+i\xi {\left|W\right|}^2\right).$$

Proof. 

The proof now follows by Theorem 5 by taking $A={\left|S^{*}\right|}^2,$ $B={\left|V\right|}^2,$ $C=SV,$ $D={\left|T^{*}\right|}^2,$ $E={\left|W\right|}^2$ and $F=TW.$ □

Now, let us move on to some applications for the pair $\left(A,A^{*}\right)$. We note that for $r\ge 1$, with $B=A$ and $C=A^{*}$, we have

$$w_r\left(A,A^{*}\right):={\displaystyle \underset{∥x∥=1}{sup}}{\left({〈\left|x,Ax\right|〉}^r+{\left|〈x,A^{*}x〉\right|}^r\right)}^{\frac{1}{r}}=2^{\frac{1}{r}}w\left(A\right).$$

By (5), we then obtain for $\lambda ,\mu ,\nu ,\xi \in \mathbb{C}$ and $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$ and such that ${\left|\lambda \right|}^q+{\left|\mu \right|}^q={\left|\nu \right|}^q+{\left|\xi \right|}^q=1,$ that

$$\begin{array}{cc}\hfill w^2\left(A\right)& \ge {\displaystyle \frac{1}{2^{\frac{2+p}{p}}}}w\left[\left({\lambda }^2+{\nu }^2\right)A^2+\left(\lambda \mu +\nu \xi \right)\left({\left|A^{*}\right|}^2+{\left|A\right|}^2\right)+\left({\mu }^2+{\xi }^2\right){\left(A^{*}\right)}^2\right]\hfill \\ \hfill & +{\displaystyle \frac{1}{2^{\frac{2+p}{p}}}}w\left[\left(\lambda +\nu \right)A+\left(\mu +\xi \right)A^{*}\right]\left|w\left(\lambda A+\mu A^{*}\right)-w\left(\nu A+\xi A^{*}\right)\right|.\hfill \end{array}$$

If $\lambda ,\mu ,\nu ,\xi \in \mathbb{C}$ with ${\left|\lambda \right|}^2+{\left|\mu \right|}^2={\left|\nu \right|}^2+{\left|\xi \right|}^2=1,$ then applying Corollary 1 with $B=A$ and $C=A^{*}$, we obtain

$$\begin{array}{cc}\hfill w^2\left(A\right)& \ge {\displaystyle \frac{1}{4}}w\left[\left({\lambda }^2+{\nu }^2\right)A^2+\left(\lambda \mu +\nu \xi \right)\left({\left|A^{*}\right|}^2+{\left|A\right|}^2\right)+\left({\mu }^2+{\xi }^2\right){\left(A^{*}\right)}^2\right]\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}w\left[\left(\left(\lambda +\nu \right)A+\left(\mu +\xi \right)A^{*}\right)\right]\left|w\left[\left(\lambda A+\mu A^{*}\right)\right]-w\left[\left(\nu A+\xi A^{*}\right)\right]\right|\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill w^2\left(A\right)& \ge {\displaystyle \frac{1}{8}}w\left\{\left[{\left(\lambda +\mu \right)}^2+{\left(\nu +\xi \right)}^2\right]A^2+\left[{\left(\lambda -\mu \right)}^2+{\left(\nu -\xi \right)}^2\right]{\left(A^{*}\right)}^2\right.\hfill \\ \hfill & \left.+\left({\lambda }^2+{\nu }^2-{\mu }^2-{\xi }^2\right)\left({\left|A^{*}\right|}^2+{\left|A\right|}^2\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{8}}w\left[\left(\lambda +\nu +\mu +\xi \right)A+\left(\lambda +\nu -\mu -\xi \right)A^{*}\right]\hfill \\ \hfill & \times \left|w\left[\left(\lambda +\mu \right)A+\left(\lambda -\mu \right)A^{*}\right]-w\left[\left(\nu +\xi \right)A+\left(\nu -\xi \right)A^{*}\right]\right|.\hfill \end{array}$$

Let $\lambda ,\mu \in \mathbb{C}$ and $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$ and such that ${\left|\lambda \right|}^q+{\left|\mu \right|}^q=1.$ By (14), we obtain for $B=A$, $C=A^{*}$ that

$$w^2\left(A\right)\ge {\displaystyle \frac{1}{2^{\frac{2}{p}}}}w\left[{\lambda }^2{A}^2+\lambda \mu \left({\left|A^{*}\right|}^2+{\left|A\right|}^2\right)+{\mu }^2{\left(A^{*}\right)}^2\right]$$

while by (17),

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

If $\lambda ,\mu \in \mathbb{C}$ and ${\left|\lambda \right|}^2+{\left|\mu \right|}^2=1$, we obtain from (18) and (19) that

$$\begin{array}{cc}\hfill w^2\left(A\right)& \ge {\displaystyle \frac{1}{4}}w\left[\left({\lambda }^2+{\mu }^2\right)A^2+2\lambda \mu \left({\left|A^{*}\right|}^2+{\left|A\right|}^2\right)+\left({\mu }^2+{\lambda }^2\right){\left(A^{*}\right)}^2\right]\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}\left|\lambda +\mu \right|∥A+A^{*}∥\left|w\left(\lambda A+\mu A^{*}\right)-w\left(\mu A+\lambda A^{*}\right)\right|\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill w^2\left(A\right)& \ge {\displaystyle \frac{1}{4}}w\left[{\left(\lambda +\mu \right)}^2{A}^2+{\left(\lambda -\mu \right)}^2{\left(A^{*}\right)}^2\right]\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}\left|\lambda +\mu \right|w\left(A\right)\hfill \\ \hfill & \times \left|w\left[\left(\lambda +\mu \right)A+\left(\lambda -\mu \right)A^{*}\right]-w\left[\left(\lambda +\mu \right)A-\left(\lambda -\mu \right)A^{*}\right]\right|.\hfill \end{array}$$

Also, if ${\left|\lambda \right|}^2+{\left|\mu \right|}^2=1,$ then by (20) and (21),

$$\begin{array}{cc}\hfill w^2\left(A\right)& \ge {\displaystyle \frac{1}{4}}\left|{\lambda }^2-{\mu }^2\right|∥A^2-{\left(A^{*}\right)}^2∥\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}w\left[\left(\lambda +i\mu \right)A+\left(\mu +i\lambda \right)A^{*}\right]\left|w\left(\lambda A+\mu A^{*}\right)-w\left(\mu A+\lambda A^{*}\right)\right|\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill w^2\left(A\right)& \ge {\displaystyle \frac{1}{8}}\left|{\lambda }^2-{\mu }^2\right|∥{\left|A^{*}\right|}^2+{\left|A\right|}^2∥\hfill \\ \hfill & +{\displaystyle \frac{1}{8}}w\left[\left(1+i\right)\left(\lambda +\mu \right)A+\left(1-i\right)\left(\lambda -\mu \right)A^{*}\right]\hfill \\ \hfill & \times \left|w\left[\left(\lambda +\mu \right)A+\left(\lambda -\mu \right)A^{*}\right]-w\left[\left(\lambda +\mu \right)A-\left(\lambda -\mu \right)A^{*}\right]\right|.\hfill \end{array}$$

From (27), we obtain

$$\begin{array}{cc}\hfill w^2\left(A\right)& \ge {\displaystyle \frac{1}{8}}∥{\left|A^{*}\right|}^2+{\left|A\right|}^2∥+{\displaystyle \frac{1}{8}}∥\left(1+i\right)A+\left(1-i\right)A^{*}∥\left|∥A+A^{*}∥-∥A-A^{*}∥\right|,\hfill \end{array}$$

while from (29)

$$\begin{array}{cc}\hfill w^2\left(A\right)& \ge {\displaystyle \frac{1}{8}}∥{\left|A^{*}\right|}^2+{\left|A\right|}^2∥+{\displaystyle \frac{\sqrt{2}}{8}}∥A+A^{*}∥\left|∥A+i{A}^{*}∥-∥A-i{A}^{*}∥\right|\hfill \end{array}$$

since the operators $\left(1+i\right)A+\left(1-i\right)A^{*}$, $A-A^{*}$, $A+i{A}^{*}$ and $A-i{A}^{*}$ are normal operators.

For any $A\in \mathcal{L}\left(H\right)$, $\varphi \in \left[0,1\right]$, and $p\ge 1$, we have by Theorem 2 with $B=A$ and $C=A^{*}$ that

$$\begin{array}{cc}\hfill w^p\left(A\right)& \le {\displaystyle \frac{1}{2}}{w}_p^{\frac{p}{2}}\left({\left|A\right|}^{2\varphi },{\left|A^{*}\right|}^{2\varphi }\right)w_p^{\frac{p}{2}}\left({\left|A^{*}\right|}^{2\left(1-\varphi \right)},{\left|A\right|}^{2\left(1-\varphi \right)}\right)\hfill \\ \hfill & \le {∥{\left|A\right|}^{2\varphi p}+{\left|A^{*}\right|}^{2\varphi p}∥}^{\frac{1}{2}}{∥{\left|A^{*}\right|}^{2\left(1-\varphi \right)p}+{\left|A\right|}^{2\left(1-\varphi \right)p}∥}^{\frac{1}{2}}\hfill \end{array}$$

and

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

The case $\varphi ={\displaystyle \frac{1}{2}}$ gives for $p\ge 1$ that

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

and

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

By Theorem 3, for any $A\in \mathcal{L}\left(H\right)$, $\varphi \in \left[0,1\right]$, $p\ge 2$, and $\lambda ,\mu ,\nu ,\xi \in \left\{-1,1\right\}$, we have

$$w^p\left(A\right)\le {\displaystyle \frac{1}{2}}w\left(\lambda {\left|A\right|}^{\varphi p}+i\mu {\left|A^{*}\right|}^{\varphi p}\right)w\left(\nu {\left|A^{*}\right|}^{\left(1-\varphi \right)p}+i\xi {\left|A\right|}^{\left(1-\varphi \right)p}\right).$$

In particular, for $\varphi ={\displaystyle \frac{1}{2}}$,

$$w^p\left(A\right)\le {\displaystyle \frac{1}{2}}w\left(\lambda {\left|A\right|}^{\frac{p}{2}}+i\mu {\left|A^{*}\right|}^{\frac{p}{2}}\right)w\left(\nu {\left|A^{*}\right|}^{\frac{p}{2}}+i\xi {\left|A\right|}^{\frac{p}{2}}\right).$$

From Corollary 3 for $S=V$ and $T=W$, we obtain

$$\begin{array}{cc}\hfill w_p^p\left(V^2,T^2\right)& \le w_p^{\frac{p}{2}}\left({\left|V^{*}\right|}^2,{\left|T^{*}\right|}^2\right)w_p^{\frac{p}{2}}\left({\left|V\right|}^2,{\left|T\right|}^2\right)\hfill \\ \hfill & \le {∥{\left|V^{*}\right|}^{2p}+{\left|T^{*}\right|}^{2p}∥}^{\frac{1}{2}}{∥{\left|V\right|}^{2p}+{\left|T\right|}^{2p}∥}^{\frac{1}{2}}\hfill \end{array}$$

and

$$w_p^p\left(V^2,T^2\right)\le {\displaystyle \frac{1}{2}}∥{\left|V\right|}^{2p}+{\left|T\right|}^{2p}+{\left|V^{*}\right|}^{2p}+{\left|T^{*}\right|}^{2p}∥.$$

Now, if we take $V=A$ and $T=A^{*}$, we obtain

$$w_p^p\left(A^2,{\left(A^{*}\right)}^2\right)\le w_p^p\left({\left|A^{*}\right|}^2,{\left|A\right|}^2\right)\le ∥{\left|A^{*}\right|}^{2p}+{\left|A\right|}^{2p}∥.$$

From Corollary 3 we obtain for $T,V\in \mathcal{L}\left(H\right)$ and $p\ge 2$ that

$$w_p^p\left(V^2,T^2\right)\le w\left(\lambda {\left|V^{*}\right|}^p+i\mu {\left|T^{*}\right|}^p\right)w\left(\nu {\left|V\right|}^p+i\xi {\left|T\right|}^p\right),$$

where $p\ge 2$ and $\lambda ,\mu ,\nu ,\xi \in \left\{-1,1\right\}.$

Now, if we take $V=A$ and $T=A^{*},$ then

$$w_p^p\left(A^2,{\left(A^{*}\right)}^2\right)\le w\left(\lambda {\left|A^{*}\right|}^p+i\mu {\left|A\right|}^p\right)w\left(\nu {\left|A\right|}^p+i\xi {\left|A^{*}\right|}^p\right),$$

where $p\ge 2$ and $\lambda ,\mu ,\nu ,\xi \in \left\{-1,1\right\}.$

Our focus now shifts to presenting some applications for the Davis–Wielandt p-radius.

Let $\lambda ,\mu ,\nu ,\xi \in \mathbb{C}$ and $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$ and such that ${\left|\lambda \right|}^q+{\left|\mu \right|}^q={\left|\nu \right|}^q+{\left|\xi \right|}^q=1.$ Then by (5) for $\left(B,C\right)=\left(T,{\left|T\right|}^2\right)$, we obtain

$$\begin{array}{cc}\hfill & d^2{w}_p\left(T\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}w\left[\left({\lambda }^2+{\nu }^2\right)T^2+\left(\lambda \mu +\nu \xi \right)\left(T{\left|T\right|}^2+{\left|T\right|}^{2}T\right)+\left({\mu }^2+{\xi }^2\right){\left|T\right|}^4\right]\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}w\left[\left(\lambda +\nu \right)T+\left(\mu +\xi \right){\left|T\right|}^2\right]\left|w\left(\lambda T+\mu {\left|T\right|}^2\right)-w\left(\nu B+\xi {\left|T\right|}^2\right)\right|.\hfill \end{array}$$

Let $\lambda ,\mu \in \mathbb{C}$ and $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$ and such that ${\left|\lambda \right|}^q+{\left|\mu \right|}^q=1.$ Then, by (14), we have

$$d^2{w}_p\left(T\right)\ge w\left[{\lambda }^2{T}^2+\lambda \mu \left(T{\left|T\right|}^2+{\left|T\right|}^{2}T\right)+{\mu }^2{\left|T\right|}^4\right]$$

and, in particular,

$$d^2{w}_p\left(T\right)\ge {\displaystyle \frac{1}{2^{2/q}}}w\left[{\left(T+{\left|T\right|}^2\right)}^2\right].$$

Let $\lambda ,\mu ,\nu ,\xi \in \mathbb{C}$ with ${\left|\lambda \right|}^2+{\left|\mu \right|}^2={\left|\nu \right|}^2+{\left|\xi \right|}^2=1.$ Then, by Corollary 1 for $\left(B,C\right)=\left(T,{\left|T\right|}^2\right)$, we obtain the following lower bounds for the Davis–Wielandt radius:

$$\begin{array}{cc}\hfill & d^{2}w\left(T\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}w\left[\left({\lambda }^2+{\nu }^2\right)T^2+\left(\lambda \mu +\nu \xi \right)\left(T{\left|T\right|}^2+{\left|T\right|}^{2}T\right)+\left({\mu }^2+{\xi }^2\right){\left|T\right|}^4\right]\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}w\left[\left(\left(\lambda +\nu \right)T+\left(\mu +\xi \right){\left|T\right|}^2\right)\right]\left|w\left[\left(\lambda T+\mu {\left|T\right|}^2\right)\right]-w\left[\left(\nu T+\xi {\left|T\right|}^2\right)\right]\right|\hfill \end{array}$$

and

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

If ${\left|\lambda \right|}^2+{\left|\mu \right|}^2=1,$ then by (15) and (16), we obtain for $\left(B,C\right)=\left(T,{\left|T\right|}^2\right)$ that

$$d^{2}w\left(T\right)\ge w\left[{\lambda }^2{T}^2+\lambda \mu \left(T{\left|T\right|}^2+{\left|T\right|}^{2}T\right)+{\mu }^2{\left|T\right|}^4\right]$$

and

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

From (26), we derive that

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

while from (28), we obtain

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

Further, we can provide the following upper bounds.

From Theorem 2 for $\left(B,C\right)=\left(T,{\left|T\right|}^2\right)$, we obtain for any $T\in \mathcal{L}\left(H\right)$, $\varphi \in \left[0,1\right]$ and $p\ge 1$ that

$$\begin{array}{cc}\hfill d^p{w}_p\left(T\right)& \le w_p^{\frac{p}{2}}\left({\left|T\right|}^{2\varphi },{\left|T\right|}^{4\varphi }\right)w_p^{\frac{p}{2}}\left({\left|T^{*}\right|}^{2\left(1-\varphi \right)},{\left|T\right|}^{4\left(1-\varphi \right)}\right)\hfill \\ \hfill & \le {∥{\left|T\right|}^{2\varphi p}+{\left|T\right|}^{4\varphi p}∥}^{\frac{1}{2}}{∥{\left|T^{*}\right|}^{2\left(1-\varphi \right)p}+{\left|T\right|}^{4\left(1-\varphi \right)p}∥}^{\frac{1}{2}}\hfill \end{array}$$

and

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

For $p=1$, we obtain

$$\begin{array}{cc}\hfill \widetilde{dw}\left(T\right)& \le w^{\frac{1}{2}}\left({\left|T\right|}^{2\varphi },{\left|T\right|}^{4\varphi }\right)w^{\frac{1}{2}}\left({\left|T^{*}\right|}^{2\left(1-\varphi \right)},{\left|T\right|}^{4\left(1-\varphi \right)}\right)\hfill \\ \hfill & \le {∥{\left|T\right|}^{2\varphi }+{\left|T\right|}^{4\varphi }∥}^{\frac{1}{2}}{∥{\left|T^{*}\right|}^{2\left(1-\varphi \right)}+{\left|T\right|}^{4\left(1-\varphi \right)}∥}^{\frac{1}{2}}\hfill \end{array}$$

and

$$\widetilde{dw}\left(T\right)\le {\displaystyle \frac{1}{2}}∥{\left|T\right|}^{2\varphi }+{\left|T\right|}^{4\varphi }+{\left|T^{*}\right|}^{2\left(1-\varphi \right)}+{\left|T\right|}^{4\left(1-\varphi \right)}∥$$

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

In particular, for $\varphi ={\displaystyle \frac{1}{2}}$, we derive

$$\widetilde{dw}\left(T\right)\le w^{\frac{1}{2}}\left(\left|T\right|,{\left|T\right|}^2\right)w^{\frac{1}{2}}\left(\left|T^{*}\right|,{\left|T\right|}^2\right)\le {∥\left|T\right|+{\left|T\right|}^2∥}^{\frac{1}{2}}{∥\left|T^{*}\right|+{\left|T\right|}^2∥}^{\frac{1}{2}}$$

and

$$\widetilde{dw}\left(T\right)\le ∥{\displaystyle \frac{\left|T\right|+\left|T^{*}\right|}{2}}+{\left|T\right|}^2∥.$$

For $p=2$, we have the following upper bounds for the Davis–Wielandt radius

$$\begin{array}{cc}\hfill d^{2}w\left(T\right)& \le w_e\left({\left|T\right|}^{2\varphi },{\left|T\right|}^{4\varphi }\right)w_e\left({\left|T^{*}\right|}^{2\left(1-\varphi \right)},{\left|T\right|}^{4\left(1-\varphi \right)}\right)\hfill \\ \hfill & \le {∥{\left|T\right|}^{4\varphi }+{\left|T\right|}^{8\varphi }∥}^{\frac{1}{2}}{∥{\left|T^{*}\right|}^{4\left(1-\varphi \right)}+{\left|T\right|}^{8\left(1-\varphi \right)}∥}^{\frac{1}{2}}\hfill \end{array}$$

and

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

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

In particular, for $\varphi ={\displaystyle \frac{1}{2}}$, we obtain

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

and

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

From Theorem 3, for any $T\in \mathcal{L}\left(H\right)$ and $\varphi \in [0,1]$, we have, for $p\ge 2$ and $\lambda ,\mu ,\nu ,\xi \in \{-1,1\}$, that

$$d^p{w}_p\left(T\right)\le w\left(\lambda {\left|T\right|}^{\varphi p}+i\mu {\left|T\right|}^{2\varphi p}\right)w\left(\nu {\left|T^{*}\right|}^{\left(1-\varphi \right)p}+i\xi {\left|T\right|}^{2\left(1-\varphi \right)p}\right),$$

which for $\varphi ={\displaystyle \frac{1}{2}}$ gives

$$d^p{w}_p\left(T\right)\le w\left(\lambda {\left|T\right|}^{\frac{p}{2}}+i\mu {\left|T\right|}^p\right)w\left(\nu {\left|T^{*}\right|}^{\frac{p}{2}}+i\xi {\left|T\right|}^p\right).$$

For $p=2$, we obtain

$$d^2{w}_e\left(T\right)\le w\left(\lambda {\left|T\right|}^{2\varphi }+i\mu {\left|T\right|}^{4\varphi }\right)w\left(\nu {\left|T^{*}\right|}^{2\left(1-\varphi \right)}+i\xi {\left|T\right|}^{4\left(1-\varphi \right)}\right)$$

and in particular,

$$d^2{w}_e\left(T\right)\le w\left(\lambda \left|T\right|+i\mu {\left|T\right|}^2\right)w\left(\nu \left|T^{*}\right|+i\xi {\left|T\right|}^2\right).$$

We conclude this paper with some applications for the pair $\left(\mathfrak{R}\left(A\right),\mathfrak{I}\left(A\right)\right)$.

Let $A\in \mathcal{L}\left(H\right)$ and $r\ge 1$. For the pair $\left(B,C\right)=\left(\mathfrak{R}\left(A\right),\mathfrak{I}\left(A\right)\right)$, we consider the functional

$${\mu }_r\left(A\right):=w_r\left(\mathfrak{R}\left(A\right),\mathfrak{I}\left(A\right)\right).$$

Observe that ${\mu }_r\left(A\right)\ge 0$ and ${\mu }_r\left(A\right)=0$ if and only if $A=0$.

Also, by the properties of the r-spherical numerical radius, we have for $T,V\in \mathcal{L}\left(H\right)$ that

$$\begin{array}{cc}\hfill {\mu }_r\left(T+V\right)& :=w_r\left(\mathfrak{R}(T+V),\mathfrak{I}(T+V)\right)\hfill \\ \hfill & =w_r\left(\mathfrak{R}\left(T\right)+\mathfrak{R}\left(V\right),\mathfrak{I}\left(T\right)+\mathfrak{I}\left(V\right)\right)\hfill \\ \hfill & =w_r\left(\left(\mathfrak{R}\left(T\right),\mathfrak{I}\left(T\right)\right)+\left(\mathfrak{R}\left(V\right),\mathfrak{I}\left(V\right)\right)\right)\hfill \\ \hfill & \le w_r\left(\left(\mathfrak{R}\left(T\right),\mathfrak{I}\left(T\right)\right)\right)+w_r\left(\left(\mathfrak{R}\left(V\right),\mathfrak{I}\left(V\right)\right)\right)\hfill \\ \hfill & ={\mu }_r\left(T\right)+{\mu }_r\left(V\right)\hfill \end{array}$$

and for $\lambda \in \mathbb{R}$, we have

$$\begin{array}{cc}\hfill {\mu }_r\left(\lambda T\right)& =w_r\left(\mathfrak{R}\left(\lambda T\right),\mathfrak{I}\left(\lambda T\right)\right)=w_r\left(\lambda \left(\mathfrak{R}\left(T\right),\mathfrak{I}\left(T\right)\right)\right)\hfill \\ \hfill & =\left|\lambda \right|w_r\left(\mathfrak{R}\left(T\right),\mathfrak{I}\left(T\right)\right),\hfill \end{array}$$

which shows that ${\mu }_r\left(·\right)$ is a norm on $\mathcal{L}\left(H\right).$

For $r=1$, we put

$$\mu \left(A\right):={\mu }_1\left(A\right)={\displaystyle \underset{∥x∥=1}{sup}}\left(\left|〈x,\mathfrak{R}\left(A\right)x〉\right|+\left|〈x,\mathfrak{I}\left(A\right)x〉\right|\right).$$

For $r=2$, we observe that

$$\begin{array}{cc}\hfill {\mu }_2\left(A\right)& :=w_e\left(\mathfrak{R}\left(A\right),\mathfrak{I}\left(A\right)\right)={\displaystyle \underset{∥x∥=1}{sup}}{\left({\left|〈x,\mathfrak{R}\left(A\right)x〉\right|}^2+{\left|〈x,\mathfrak{I}\left(A\right)x〉\right|}^2\right)}^{\frac{1}{2}}\hfill \\ \hfill & ={\displaystyle \underset{∥x∥=1}{sup}}\left|〈Ax,x〉\right|=w\left(A\right).\hfill \end{array}$$

Let $\lambda ,\mu ,\nu ,\xi \in \mathbb{C}$ and $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$ and such that ${\left|\lambda \right|}^q+{\left|\mu \right|}^q={\left|\nu \right|}^q+{\left|\xi \right|}^q=1.$ Then, by (5) for $\left(B,C\right)=\left(\mathfrak{R}\left(A\right),\mathfrak{I}\left(A\right)\right)$, we have

$$\begin{array}{cc}\hfill {\mu }_p^2\left(A\right)& \ge {\displaystyle \frac{1}{2}}w\left[\left({\lambda }^2+{\nu }^2\right){\mathfrak{R}}^2\left(A\right)+\left({\mu }^2+{\xi }^2\right){\mathfrak{I}}^2\left(A\right)\right.\hfill \\ \hfill & \left.+\left(\lambda \mu +\nu \xi \right)\left(\mathfrak{R}\left(A\right)\mathfrak{I}\left(A\right)+\mathfrak{I}\left(A\right)\mathfrak{R}\left(A\right)\right)\right]\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}w\left[\left(\lambda +\nu \right)\mathfrak{R}\left(A\right)+\left(\mu +\xi \right)\mathfrak{I}\left(A\right)\right]\hfill \\ \hfill & \times \left|w\left(\lambda \mathfrak{R}\left(A\right)+\mu \mathfrak{I}\left(A\right)\right)-w\left(\nu \mathfrak{R}\left(A\right)+\xi \mathfrak{I}\left(A\right)\right)\right|.\hfill \end{array}$$

Let $\lambda ,\mu \in \mathbb{C}$ and $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$ and such that ${\left|\lambda \right|}^q+{\left|\mu \right|}^q=1.$ By (14), we obtain

$${\mu }_p^2\left(A\right)\ge w\left[{\lambda }^2{\mathfrak{R}}^2\left(A\right)+\lambda \mu \left(\mathfrak{R}\left(A\right)\mathfrak{I}\left(A\right)+\mathfrak{I}\left(A\right)\mathfrak{R}\left(A\right)\right)+{\mu }^2{\mathfrak{I}}^2\left(A\right)\right]$$

and, in particular for $\lambda =\mu ={\displaystyle \frac{1}{2^{\frac{1}{q}}}},$

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

If ${\left|\lambda \right|}^q+{\left|\mu \right|}^q=1$, then by (17), we obtain

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

Let $\lambda ,\mu ,\nu ,\xi \in \mathbb{C}$ with ${\left|\lambda \right|}^2+{\left|\mu \right|}^2={\left|\nu \right|}^2+{\left|\xi \right|}^2=1.$ Then, by Corollary 1 with $(B,C)=\left(\mathfrak{R}\left(A\right),\mathfrak{I}\left(A\right)\right)$, we obtain

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

and

$$\begin{array}{cc}\hfill & w^2\left(A\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{4}}w\left\{\left[{\left(\lambda +\mu \right)}^2+{\left(\nu +\xi \right)}^2\right]{\mathfrak{R}}^2\left(A\right)+\left[{\left(\lambda -\mu \right)}^2+{\left(\nu -\xi \right)}^2\right]{\mathfrak{I}}^2\left(A\right)\right.\hfill \\ \hfill & \left.+\left({\lambda }^2+{\nu }^2-{\mu }^2-{\xi }^2\right)\left(\mathfrak{R}\left(A\right)\mathfrak{I}\left(A\right)+\mathfrak{I}\left(A\right)\mathfrak{R}\left(A\right)\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}w\left[\left(\lambda +\nu +\mu +\xi \right)\mathfrak{R}\left(A\right)+\left(\lambda +\nu -\mu -\xi \right)\mathfrak{I}\left(A\right)\right]\hfill \\ \hfill & \times \left|w\left[\left(\lambda +\mu \right)\mathfrak{R}\left(A\right)+\left(\lambda -\mu \right)\mathfrak{I}\left(A\right)\right]-w\left[\left(\nu +\xi \right)\mathfrak{R}\left(A\right)+\left(\nu -\xi \right)\mathfrak{I}\left(A\right)\right]\right|.\hfill \end{array}$$

If ${\left|\lambda \right|}^2+{\left|\mu \right|}^2=1,$ then by (15) and (16), we also have

$$w^2\left(A\right)\ge w\left[{\lambda }^2{\mathfrak{R}}^2\left(A\right)+\lambda \mu \left(\mathfrak{R}\left(A\right)\mathfrak{I}\left(A\right)+\mathfrak{I}\left(A\right)\mathfrak{R}\left(A\right)\right)+{\mu }^2{\mathfrak{I}}^2\left(A\right)\right]$$

and

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

Also, by (18) and (19), we obtain

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

and

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

From (26), we obtain, by taking $(B,C)=\left(\mathfrak{R}\right(A),\mathfrak{I}(A\left)\right)$, that

$$w^2\left(A\right)\ge {\displaystyle \frac{1}{2}}\parallel {\mathfrak{R}}^2\left(A\right)-{\mathfrak{I}}^2\left(A\right)\parallel +{\displaystyle \frac{1}{2}}w\left(A\right)|\parallel \mathfrak{R}\left(A\right)\parallel -\parallel \mathfrak{I}\left(A\right)\parallel |$$

while, from (28), we have

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

We determine for $\varphi \in \left[0,1\right]$ and $p\ge 1$ from Theorem 2 that

$$\begin{array}{cc}\hfill {\mu }_p^p\left(A\right)& \le w_p^{\frac{p}{2}}\left({\left|\mathfrak{R}\left(A\right)\right|\varphi }^{2\varphi },{\left|\mathfrak{I}\left(A\right)\right|\varphi }^{2 }\right)w_p^{\frac{p}{2}}\left({\left|\mathfrak{R}\left(A\right)\right| }^{2\left(1-\varphi \right)},{\left|\mathfrak{I}\left(A\right)\right|}^{2\left(1-\varphi \right)}\right)\hfill \\ \hfill & \le {∥{\left|\mathfrak{R}\left(A\right)\right| }^{2\varphi p}+{\left|\mathfrak{I}\left(A\right)\right|}^{2\varphi p}∥}^{\frac{1}{2}}{∥{\left|\mathfrak{R}\left(A\right)\right| }^{2\left(1-\varphi \right)p}+{\left|\mathfrak{I}\left(A\right)\right|}^{2\left(1-\varphi \right)p}∥}^{\frac{1}{2}}\hfill \end{array}$$

and

$${\mu }_p^p\left(A\right)\le {\displaystyle \frac{1}{2}}∥{\left|\mathfrak{R}\left(A\right)\right| }^{2\varphi p}+{\left|\mathfrak{I}\left(A\right)\right| }^{2\varphi p}+{\left|\mathfrak{R}\left(A\right)\right| }^{2\left(1-\varphi \right)p}+{\left|\mathfrak{I}\left(A\right)\right| }^{2\left(1-\varphi \right)p}∥.$$

For $p=1$, we obtain

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

and

$$\mu \left(A\right)\le {\displaystyle \frac{1}{2}}∥{\left|\mathfrak{R}\left(A\right)\right| }^{2\varphi }+{\left|\mathfrak{I}\left(A\right)\right|}^{2\varphi }+{\left|\mathfrak{R}\left(A\right)\right| }^{2\left(1-\varphi \right)}+{\left|\mathfrak{I}\left(A\right)\right|}^{2\left(1-\varphi \right)}∥$$

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

If we take $\varphi ={\displaystyle \frac{1}{2}},$ then we have

$$\mu \left(A\right)\le w^{\frac{1}{2}}\left(\left|\mathfrak{R}\left(A\right)\right|,\left|\mathfrak{I}\left(A\right)\right|\right)\le ∥\left|\mathfrak{R}\left(A\right)\right|+\left|\mathfrak{I}\left(A\right)\right|∥.$$

For $p=2$, we obtain

$$\begin{array}{cc}\hfill w^2\left(A\right)& \le w_e\left({\left|\mathfrak{R}\left(A\right)\right| }^{2\varphi },{\left|\mathfrak{I}\left(A\right)\right|}^{2\varphi }\right)w_e\left({\left|\mathfrak{R}\left(A\right)\right| }^{2\left(1-\varphi \right)},{\left|\mathfrak{I}\left(A\right)\right|}^{2\left(1-\varphi \right)}\right)\hfill \\ \hfill & \le {∥{\left|\mathfrak{R}\left(A\right)\right| }^{4\varphi }+{\left|\mathfrak{I}\left(A\right)\right|}^{4\varphi }∥}^{\frac{1}{2}}{∥{\left|\mathfrak{R}\left(A\right)\right| }^{4\left(1-\varphi \right)}+{\left|\mathfrak{I}\left(A\right)\right|}^{4\left(1-\varphi \right)}∥}^{\frac{1}{2}}\hfill \end{array}$$

and

$$w^2\left(A\right)\le {\displaystyle \frac{1}{2}}∥{\left|\mathfrak{R}\left(A\right)\right| }^{4\varphi }+{\left|\mathfrak{I}\left(A\right)\right|}^{4\varphi }+{\left|\mathfrak{R}\left(A\right)\right| }^{4\left(1-\varphi \right)}+{\left|\mathfrak{I}\left(A\right)\right|}^{4\left(1-\varphi \right)}∥$$

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

If we take $\varphi ={\displaystyle \frac{1}{2}},$ then we obtain

$$w\left(A\right)\le w_e\left(\left|\mathfrak{R}\left(A\right)\right|,\left|\mathfrak{I}\left(A\right)\right|\right)\le {∥{\left|\mathfrak{R}\left(A\right)\right| }^2+{\left|\mathfrak{I}\left(A\right)\right|}^2∥}^{\frac{1}{2}}.$$

From Theorem 3 we obtain for $\varphi \in \left[0,1\right]$, $p\ge 2$ and $\lambda ,\mu ,\nu ,\xi \in \left\{-1,1\right\}$ that

$${\mu }_p^p\left(A\right)\le w\left(\lambda {\left|\mathfrak{R}\left(A\right)\right| }^{\varphi p}+i\mu {\left|\mathfrak{I}\left(A\right)\right|}^{\varphi p}\right)w\left(\nu {\left|\mathfrak{R}\left(A\right)\right| }^{\left(1-\varphi \right)p}+i\xi {\left|\mathfrak{I}\left(A\right)\right|}^{\left(1-\varphi \right)p}\right).$$

For $p=1$, we derive that

$$\mu \left(A\right)\le w\left(\lambda {\left|\mathfrak{R}\left(A\right)\right| }^{\varphi }+i\mu {\left|\mathfrak{I}\left(A\right)\right|}^{\varphi }\right)w\left(\nu {\left|\mathfrak{R}\left(A\right)\right| }^{1-\varphi }+i\xi {\left|\mathfrak{I}\left(A\right)\right|}^{1-\varphi }\right)$$

for $\varphi \in \left[0,1\right],$ which gives for $\varphi ={\displaystyle \frac{1}{2}}$ that

$$\mu \left(A\right)\le w\left(\lambda {\left|\mathfrak{R}\left(A\right)\right| }^{\frac{1}{2}}+i\mu {\left|\mathfrak{I}\left(A\right)\right|}^{\frac{1}{2}}\right)w\left(\nu {\left|\mathfrak{R}\left(A\right)\right|}^{\frac{1}{2}}+i\xi {\left|\mathfrak{I}\left(A\right)\right|}^{\frac{1}{2}}\right).$$

For $p=2$, we obtain

$$\begin{array}{cc}\hfill w^2\left(A\right)& \le w\left(\lambda {\left|\mathfrak{R}\left(A\right)\right| }^{2\varphi }+i\mu {\left|\mathfrak{I}\left(A\right)\right| }^{2\varphi }\right)\hfill \\ \hfill & \times w\left(\nu {\left|\mathfrak{R}\left(A\right)\right| }^{2\left(1-\varphi \right)}+i\xi {\left|\mathfrak{I}\left(A\right)\right|}^{2\left(1-\varphi \right)}\right)\hfill \end{array}$$

for $\varphi \in \left[0,1\right],$ which gives for $\varphi ={\displaystyle \frac{1}{2}}$ that

$$w^2\left(A\right)\le w\left(\lambda \left|\mathfrak{R}\left(A\right)\right|+i\mu \left|\mathfrak{I}\left(A\right)\right|\right)w\left(\nu \left|\mathfrak{R}\left(A\right)\right|+i\xi \left|\mathfrak{I}\left(A\right)\right|\right).$$

3. Conclusions

As we have seen, the authors have shown the lower and upper bounds for the generalized r–numerical radius $w_r(B,C)$ for a pair of the bounded linear operators B and C. It is natural to consider the bounds of the following r–numerical radius $w_r(A_1,\cdots ,A_n)$ introduced in [22] for n–tuples of the bounded linear operators $A_1,\cdots ,A_n$:

$$w_r(A_1,\cdots ,A_n):={\displaystyle \underset{\parallel x\parallel =1}{sup}}{\left({\displaystyle \sum _{k=1}^n}{\left|\langle A_{k}x,x\rangle \right|}^r\right)}^{1/r},(r\ge 1).$$

Then we can obtain the following bounds of $w_r(A_1,\cdots ,A_n)$ by the usual numerical radius by using a proof similar to that of Theorems 1 and 2.

Proposition 1.

Let $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$ and ${\alpha }_1^{\left(l\right)},\cdots ,{\alpha }_n^{\left(l\right)}\in \mathbb{C}$ such that ${\sum }_{k=1}^n{\left|{\alpha }_k^{\left(l\right)}\right|}^q=1$ for any $l=1,2,\cdots ,n$. For any $l=1,2,\cdots ,n$ and $A_1,\cdots ,A_n\in \mathcal{L}\left(H\right)$,

$$w_p(A_1,\cdots ,A_n)\ge w\left({\displaystyle \sum _{k=1}^n}{\alpha }_k^{\left(l\right)}{A}_k\right).$$

Proof. 

For any $l=1,2,\cdots ,n$, by using the same method as in the proof of Theorem 1,

$$\begin{array}{cc}\hfill {\left({\displaystyle \sum _{k=1}^n}{\left|\langle A_{k}x,x\rangle \right|}^p\right)}^{1/p}& ={\left({\displaystyle \sum _{k=1}^n}{\left|{\alpha }_k^{\left(l\right)}\right|}^q\right)}^{1/q}{\left({\displaystyle \sum _{k=1}^n}{\left|\langle A_{k}x,x\rangle \right|}^p\right)}^{1/p}\hfill \\ \hfill & \ge \left|{\displaystyle \sum _{k=1}^n}{\alpha }_k^{\left(l\right)}\langle A_{k}x,x\rangle \right|=\left|〈{\displaystyle \sum _{k=1}^n}{\alpha }_k^{\left(l\right)}{A}_{k}x,x〉\right|.\hfill \end{array}$$

Taking the supremum over $x\in H$ with $\parallel x\parallel =1$, we obtain the result. □

Proposition 2.

Let $p\ge 1$ and $\varphi \in [0,1]$. For any $A_1,\cdots ,A_n\in \mathcal{L}\left(H\right)$,

$$w_p(A_1,\cdots ,A_n)\le w^{1/p}\left({\displaystyle \sum _{k=1}^n}\left({\displaystyle \frac{|A_k{|}^{2\varphi p}+{\left|A_k^{*}\right|}^{2(1-\varphi )p}}{2}}\right)\right).$$

Proof. 

We calculate by the same method as that used in the proof of Theorem 2,

$$\begin{array}{cc}\hfill {\displaystyle \sum _{k=1}^n}{\left|\langle A_{k}x,x\rangle \right|}^p& \le {\displaystyle \sum _{k=1}^n}\langle |A_k{|}^{2\varphi }{x,x\rangle }^{p/2}\langle |A_k^{*}{{|}^{2(1-\varphi )p}x,x\rangle }^{p/2}\hfill \\ \hfill & \le {\left({\displaystyle \sum _{k=1}^n}\langle |A_k{{|}^{2\varphi }x,x\rangle }^p\right)}^{1/2}{\left({\displaystyle \sum _{k=1}^n}\langle |A_k^{*}{{|}^{2(1-\varphi )}x,x\rangle }^p\right)}^{1/2}\hfill \\ \hfill & \le {\left({\displaystyle \sum _{k=1}^n}\langle |A_k{|}^{2\varphi p}x,x\rangle \right)}^{1/2}{\left({\displaystyle \sum _{k=1}^n}\langle |A_k^{*}{|}^{2(1-\varphi )p}x,x\rangle \right)}^{1/2}\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left({\displaystyle \sum _{k=1}^n}\langle |A_k{|}^{2\varphi p}x,x\rangle +{\displaystyle \sum _{k=1}^n}\langle |A_k^{*}{|}^{2(1-\varphi )p}x,x\rangle \right)\hfill \\ \hfill & =〈{\displaystyle \sum _{k=1}^n}\left({\displaystyle \frac{|A_k{|}^{2\varphi p}+{|A_k^{*}|}^{2(1-\varphi )p}}{2}}\right)x,x〉.\hfill \end{array}$$

Taking the power $1/p$ on both sides and the supremum over $x\in H$ with $\parallel x\parallel =1$, we obtain the result. □

We consider the case $0<p\le 1$. To this end, we review that the following inequalities hold (e.g., [25] [Theorem 2.11]):

$$K(m,M,p){\langle Ax,x\rangle }^p\le \langle A^{p}x,x\rangle \le {\langle Ax,x\rangle }^p,(0\le p\le 1)$$

(44)

for every unit vector $x\in H$ and a positive operator A such that $mI\le A\le MI$ for positive scalars $0<m<M$, where $K(m,M,p)$ is often called the generalized Kantorovich constant and is given by

$$K(m,M,p):={\displaystyle \frac{(m{M}^p-M{m}^p)}{(p-1)(M-m)}}{\left({\displaystyle \frac{p-1}{p}}{\displaystyle \frac{M^p-m^p}{m{M}^p-M{m}^p}}\right)}^p.(p\in \mathbb{R}).$$

Then, we can obtain the following result.

Proposition 3.

Let $0<p\le 1$ and $\varphi \in [0,1]$. For any $A_1,\cdots ,A_n\in \mathcal{L}\left(H\right)$ such that $m\le |A_k|\le M$ and $m\le |A_k^{*}|\le M$ for some scalar $0<m<M$ and all $k=1,2,\cdots ,n$. Then,

$$w_p(A_1,\cdots ,A_n)\le {\displaystyle \frac{1}{K^{1/p}(m,M,p)}}{w}^{1/p}\left({\displaystyle \sum _{k=1}^n}\left({\displaystyle \frac{|A_k{|}^{2\varphi p}+{|A_k^{*}|}^{2(1-\varphi )p}}{2}}\right)\right).$$

Proof. 

We calculate by a method similar to that used in the proof of Proposition 2,

$$\begin{array}{cc}\hfill {\displaystyle \sum _{k=1}^n}{|\langle A_{k}x,x\rangle |}^p& \le {\displaystyle \sum _{k=1}^n}\langle |A_k{|}^{2\varphi }{x,x\rangle }^{p/2}\langle |A_k^{*}{{|}^{2(1-\varphi )p}x,x\rangle }^{p/2}\hfill \\ \hfill & \le {\left({\displaystyle \sum _{k=1}^n}\langle |A_k{{|}^{2\varphi }x,x\rangle }^p\right)}^{1/2}{\left({\displaystyle \sum _{k=1}^n}\langle |A_k^{*}{{|}^{2(1-\varphi )}x,x\rangle }^p\right)}^{1/2}\hfill \\ \hfill & \le {\displaystyle \frac{1}{K(m,M,p)}}{\left({\displaystyle \sum _{k=1}^n}\langle |A_k{|}^{2\varphi p}x,x\rangle \right)}^{1/2}{\left({\displaystyle \sum _{k=1}^n}\langle |A_k^{*}{|}^{2(1-\varphi )p}x,x\rangle \right)}^{1/2}\hfill \\ \hfill & \le {\displaystyle \frac{1}{2K(m,M,p)}}\left({\displaystyle \sum _{k=1}^n}\langle |A_k{|}^{2\varphi p}x,x\rangle +{\displaystyle \sum _{k=1}^n}\langle |A_k^{*}{|}^{2(1-\varphi )p}x,x\rangle \right)\hfill \\ \hfill & ={\displaystyle \frac{1}{K(m,M,p)}}〈{\displaystyle \sum _{k=1}^n}\left({\displaystyle \frac{|A_k{|}^{2\varphi p}+{|A_k^{*}|}^{2(1-\varphi )p}}{2}}\right)x,x〉.\hfill \end{array}$$

In the third inequality above, we used the first inequality in (44). Taking the power $1/p$ on both sides and the supremum over $x\in H$ with $\parallel x\parallel =1$, we obtain the result. □

From Proposition 2, we see that

$$w_p(A_1,\cdots ,A_n)\le {∥{\displaystyle \sum _{k=1}^n}\left({\displaystyle \frac{|A_k{|}^{2\varphi p}+{|A_k^{*}|}^{2(1-\varphi )p}}{2}}\right)∥}^{1/p},(p\ge 1).$$

(45)

In [22], the following upper bounds have been established for $\varphi \in [0,1]$:

$$w_p(A_1,\cdots ,A_n)\le {∥{\displaystyle \sum _{k=1}^n}{\left({\displaystyle \frac{|A_k{|}^{2\varphi }+{|A_k^{*}|}^{2(1-\varphi )}}{2}}\right)}^p∥}^{1/p},(p\ge 1)$$

(46)

and

$$w_p(A_1,\cdots ,A_n)\le {∥{\displaystyle \sum _{k=1}^n}\left(\varphi |A_k{|}^p+(1-\varphi ){|A_k^{*}|}^p\right)∥}^{1/p},(p\ge 2).$$

(47)

It may be of interest to compare (45) with (46) and (47). By the convexity of the function $x^p$ for $p\ge 1$ on $[0,\infty )$, we have

$${\left({\displaystyle \frac{{\alpha }^{\varphi }+{\beta }^{1-\varphi }}{2}}\right)}^p\le {\displaystyle \frac{{\alpha }^{\varphi p}+{\beta }^{(1-\varphi )p}}{2}},(\alpha ,\beta >0,0\le \varphi \le 1).$$

From this, the inequality (46) gives a better upper bound than our inequality (45).

Also, we easily observe that our inequality (45) coincides with the inequality (47) when $\varphi ={\displaystyle \frac{1}{2}}$. In addition, there is no general ordering between ${\displaystyle \frac{{\alpha }^{2\varphi }+{\beta }^{2(1-\varphi )}}{2}}$ and $\varphi \alpha +(1-\varphi )\beta$ for $\alpha ,\beta >0$ and $0\le \varphi \le 1$. Of course, our inequality (45) and the inequality (47) have different conditions regarding p, so they should not be directly compared. However, we believe that there is no superiority or inferiority between the upper bounds in our inequality (45) and the inequality (47) for $p\ge 2$.

However, we would like to emphasize that the important point in the main results is that we precisely derive upper and lower bounds for the pair of operators.