关于与几何平均相关的范数不等式
泛函分析
2022-10-26 v1
摘要
设对所有i = 1 , ⋯ , m i=1,\cdots,m i = 1 , ⋯ , m ,A i A_i A i 和B i B_i B i 为正定矩阵。我们证明了对所有p > 0 p>0 p > 0 和所有r ≥ 1 r\geq1 r ≥ 1 ,有∣ ∣ ∑ i = 1 m ( A i 2 ♯ B i 2 ) r ∣ ∣ 1 ≤ ∣ ∣ ( ( ∑ i = 1 m A i ) p r 2 ( ∑ i = 1 m B i ) p r ( ∑ i = 1 m A i ) r p 2 ) 1 p ∣ ∣ 1 . \left|\left|\sum_{i=1}^m(A_i^2\sharp B_i^2)^r\right|\right|_1\leq\left|\left|\left(\left(\sum_{i=1}^mA_i\right)^{\frac{pr}{_2}}\left(\sum_{i=1}^mB_i\right)^{pr}\left(\sum_{i=1}^mA_i\right)^{\frac{rp}{_2}}\right)^{\frac{1}{p}}\right|\right|_1. i = 1 ∑ m ( A i 2 ♯ B i 2 ) r 1 ≤ ( i = 1 ∑ m A i ) 2 p r ( i = 1 ∑ m B i ) p r ( i = 1 ∑ m A i ) 2 r p p 1 1 . 我们猜想该不等式对所有酉不变范数也成立。我们对m = 2 m=2 m = 2 、p ≥ 1 p\geq1 p ≥ 1 、r ≥ 1 r\geq1 r ≥ 1 且对所有酉不变范数的情形给出肯定回答。换言之,我们证明了对所有酉不变范数、所有p ≥ 1 p\geq1 p ≥ 1 和所有r ≥ 1 r\geq1 r ≥ 1 ,有∣ ∣ ∣ ( A 2 ♯ B 2 ) r + ( C 2 ♯ D 2 ) r ∣ ∣ ∣ ≤ ∣ ∣ ∣ ( ( A + C ) r p 2 ( B + D ) r p ( A + C ) r p 2 ) 1 p ∣ ∣ ∣ , \left|\left|\left|\left(A^{^2}\sharp B^{^2}\right)^{r}+\left(C^{^2}\sharp D^{^2}\right)^{r}\right|\right|\right|\leq \left|\left|\left|\left(\left(A+C\right)^{^\frac{rp}{_2}}\left(B+D\right)^{{rp}}\left(A+C\right)^{^\frac{rp}{_2}}\right)^{\frac{1}{_p}}\right|\right|\right|, ( A 2 ♯ B 2 ) r + ( C 2 ♯ D 2 ) r ≤ ( ( A + C ) 2 r p ( B + D ) r p ( A + C ) 2 r p ) p 1 , 其中A , B , C , D A,B,C,D A , B , C , D 为正定矩阵。这给出了对Dinh、Ahsani和Tam所提猜想在m = 2 m=2 m = 2 情形下的肯定回答。前述不等式直接导出Audenaert最近的结果\cite{ANIFP}。
引用
@article{arxiv.2210.14023,
title = {On norm inequalities related to the geometric mean},
author = {Shaima'a Freewan and Mostafa Hayajneh},
journal= {arXiv preprint arXiv:2210.14023},
year = {2022}
}