Numerical radius inequalities of bounded linear operators and \((\alpha ,\beta )\)-normal operators

We obtain various upper bounds for the numerical radius w(T) of a bounded linear operator T defined on a complex Hilbert space \(\mathcal {H}\) $\mathcal{H}$ , by developing the upper bounds for the \(\alpha \) $\alpha$ -norm of T, which is defined as \(\Vert T\Vert _{\alpha }= \sup \left\{ \sqrt{\alpha |\langle Tx,x \rangle |^2+ (1-\alpha )\Vert Tx\Vert ^2 }: x\in \mathcal {H}, \Vert x\Vert =1 \right\} \) ${\Vert T\Vert }_{\alpha }=sup\left\{\sqrt{{\alpha |⟨Tx,x⟩|}^2+(1-\alpha ){\Vert Tx\Vert }^2},:,x,\in ,\mathcal{H},,,\Vert x\Vert ,=,1\right\}$ for \( 0\le \alpha \le 1 \) $0\le \alpha \le 1$ . Further, we prove that \(\begin{aligned} w(T)\le & \sqrt{\Big ( \left\| \alpha |T|+(1-\alpha )|T^*| \right\| \Big ) \Vert T\Vert } \,\,\,\, \le \,\, \,\, \Vert T\Vert , \,\, \forall \alpha \in [0,1]. \end{aligned}\) $$\begin{array}{cc}\hfill w(T)\le & \sqrt{(∥\alpha |T|+(1-\alpha )|,T^{\ast },|∥)\Vert T\Vert }\le \Vert T\Vert ,\forall \alpha \in [0,1].\hfill \end{array}$$ For \(0\le \alpha \le 1 \le \beta ,\) $0\le \alpha \le 1\le \beta ,$ the operator T is called \((\alpha ,\beta )\) $(\alpha ,\beta )$ -normal if \(\alpha ^2 T^*T\le TT^*\le \beta ^2 T^*T\) ${\alpha }^2{T}^{\ast }T\le T{T}^{\ast }\le {\beta }^2{T}^{\ast }T$ holds. Note that every invertible operator is an \((\alpha ,\beta )\) $(\alpha ,\beta )$ -normal operator for suitable values of \(\alpha \) $\alpha$ and \(\beta \) $\beta$ . Among other lower bounds for the numerical radius of an \((\alpha ,\beta )\) $(\alpha ,\beta )$ -normal operator T, we show that \(\begin{aligned} w(T)\ge & \sqrt{\max \left\{ 1+\alpha ^2, 1+\frac{1}{\beta ^2}\right\} \frac{\Vert T\Vert ^2}{4}+ \frac{\left| \Vert \Re (T)\Vert ^2-\Vert \Im (T)\Vert ^2 \right| }{2}} \\\ge & \max \left\{ \sqrt{1+\alpha ^2}, \sqrt{1+\frac{1}{\beta ^2}} \right\} \frac{\Vert T\Vert }{2} > \frac{\Vert T\Vert }{2}, \end{aligned}\) $$\begin{array}{cc}\hfill w(T)\ge & \sqrt{max\left\{1,+,{\alpha }^2,,,1,+,\frac{1}{{\beta }^2}\right\}\frac{{\Vert T\Vert }^2}{4}+\frac{\left|{\Vert \Re (T)\Vert }^2,-,{\Vert \Im (T)\Vert }^2\right|}{2}}\hfill \\ \hfill \ge & max\left\{\sqrt{1+{\alpha }^2},,,\sqrt{1+\frac{1}{{\beta }^2}}\right\}\frac{\Vert T\Vert }{2}>\frac{\Vert T\Vert }{2},\hfill \end{array}$$ where \(\Re (T)\) $\Re (T)$ and \(\Im (T)\) $\Im (T)$ are the real part and imaginary part of T, respectively.