当前位置: X-MOL 学术Georgian Math. J. › 论文详情
Our official English website, www.x-mol.net, welcomes your feedback! (Note: you will need to create a separate account there.)
Further refinements of generalized numerical radius inequalities for Hilbert space operators
Georgian Mathematical Journal ( IF 0.7 ) Pub Date : 2019-05-07 , DOI: 10.1515/gmj-2019-2023
Monire Hajmohamadi 1 , Rahmatollah Lashkaripour 1 , Mojtaba Bakherad 1
Affiliation  

In this paper, we show some refinements of generalized numerical radius inequalities involving the Young and Heinz inequalities. In particular, we present \begin{align*} w_{p}^{p}(A_{1}^{*}T_{1}B_{1},...,A_{n}^{*}T_{n}B_{n})\leq\frac{n^{1-\frac{1}{r}}}{2^{\frac{1}{r}}}\Big\|\sum_{i=1}^{n}[B_{i}^{*} f^{2}(|T_{i}|)B_{i}]^{rp}+[A_{i}^{*}g^{2}(|T_{i}^{*}|)A_{i}]^{rp}\Big\|^{\frac{1}{r}} -\inf_{\|x\|=1}\eta(x), \end{align*} where $T_{i}, A_{i}, B_{i} \in {\mathbb B}({\mathscr H})\,\,(1\leq i\leq n)$, $f$ and $g$ are nonnegative continuous functions on $[0, \infty)$ satisfying $f(t)g(t)=t$ for all $t\in [0, \infty)$, $p, r\geq 1$, $N\in {\mathbb N}$ and \begin{align*} \eta(x)= \frac{1}{2}\sum_{i=1}^{n}\sum_{j=1}^{N} \Big(\sqrt[2^{j}]{ \langle (A_{i}^{*}g^{2}(|T_{i}^{*}|)A_{i})^{p}x, x\rangle^{2^{j-1}-k_{j}} \langle (B_{i}^{*} f^{2}(|T_{i}|)B_{i})^{p}x, x\rangle^{k_j}}\quad-\sqrt[2^{j}]{ \langle (B_{i}^{*}f^{2}(|T_{i}|)B_{i})^{p}x, x\rangle^{k_{j}+1} \langle (A_{i}^{*} g^{2}(|T_{i}^{*}|)A_{i})^{p}x, x\rangle^{2^{j-1}-k_{j}-1}}\Big)^{2}. \end{align*}

中文翻译:

希尔伯特空间算子广义数值半径不等式的进一步改进

在本文中,我们展示了涉及 Young 和 Heinz 不等式的广义数值半径不等式的一些改进。特别地,我们提出 \begin{align*} w_{p}^{p}(A_{1}^{*}T_{1}B_{1},...,A_{n}^{*}T_ {n}B_{n})\leq\frac{n^{1-\frac{1}{r}}}{2^{\frac{1}{r}}}\Big\|\sum_{i =1}^{n}[B_{i}^{*} f^{2}(|T_{i}|)B_{i}]^{rp}+[A_{i}^{*}g^ {2}(|T_{i}^{*}|)A_{i}]^{rp}\Big\|^{\frac{1}{r}} -\inf_{\|x\|=1 }\eta(x), \end{align*} 其中 $T_{i}, A_{i}, B_{i} \in {\mathbb B}({\mathscr H})\,\,(1\ leq i\leq n)$, $f$ 和 $g$ 是 $[0, \infty)$ 上的非负连续函数,对于所有 $t\in [0, \infty)$, $p, r\geq 1$, $N\in {\mathbb N}$ 和 \begin{align*} \eta(x)= \frac{1}{2}\sum_{i= 1}^{n}\sum_{j=1}^{N} \Big(\sqrt[2^{j}]{ \langle (A_{i}^{*}g^{2}(|T_{ i}^{*}|)A_{i})^{p}x, x\rangle^{2^{j-1}-k_{j}} \langle (B_{i}^{*} f^ {2}(|T_{i}|)B_{i})^{p}x,x\rangle^{k_j}}\quad-\sqrt[2^{j}]{ \langle (B_{i}^{*}f^{2}(|T_{i}|)B_{i}) ^{p}x, x\rangle^{k_{j}+1} \langle (A_{i}^{*} g^{2}(|T_{i}^{*}|)A_{i} )^{p}x, x\rangle^{2^{j-1}-k_{j}-1}}\Big)^{2}。\end{对齐*}
更新日期:2019-05-07
down
wechat
bug