Matrix inequalities between $$f(A)\sigma f(B)$$ and $$A\sigma B$$

Pub Date : 2024-04-23 DOI:10.1007/s00010-024-01059-z
Manisha Devi, Jaspal Singh Aujla, Mohsen Kian, Mohammad Sal Moslehian
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Abstract

Let A and B be \(n\times n\) positive definite complex matrices, let \(\sigma \) be a matrix mean, and let \(f: [0,\infty )\rightarrow [0,\infty )\) be a differentiable convex function with \(f(0)=0\). We prove that

$$\begin{aligned} f^{\prime }(0)(A \sigma B)\le \frac{f(m)}{m}(A\sigma B)\le f(A)\sigma f(B)\le \frac{f(M)}{M}(A\sigma B)\le f^{\prime }(M)(A\sigma B), \end{aligned}$$

where m represents the smallest eigenvalues of A and B and M represents the largest eigenvalues of A and B. If f is differentiable and concave, then the reverse inequalities hold. We use our result to improve some known subadditivity inequalities involving unitarily invariant norms under certain mild conditions. In particular, if f(x)/x is increasing, then

$$\begin{aligned} |||f(A)+f(B)|||\le \frac{f(M)}{M} |||A+B|||\le |||f(A+B)||| \end{aligned}$$

holds for all A and B with \(M\le A+B\). Furthermore, we apply our results to explore some related inequalities. As an application, we present a generalization of Minkowski’s determinant inequality.

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$$f(A)\sigma f(B)$$ 与 $$A\sigma B$ 之间的矩阵不等式
让 A 和 B 是(n 次 n)正定复矩阵,让(\sigma \)是一个矩阵均值,让(f: [0,\infty )\rightarrow [0,\infty )\) 是一个可微凸函数,且(f(0)=0)。我们证明 $$\begin{aligned} f^{prime }(0)(A\sigma B)\le f^{f(m)}{m}(A\sigma B)\le f(A)\sigma f(B)\le f^{f(M)}{M}(A\sigma B)\le f^{prime }(M)(A\sigma B)、\end{aligned}$$其中 m 代表 A 和 B 的最小特征值,M 代表 A 和 B 的最大特征值。如果 f 是可微且凹的,则反向不等式成立。我们利用我们的结果改进了一些已知的、在某些温和条件下涉及单位不变规范的次等不等式。特别是,如果 f(x)/x 是递增的,那么 $$\begin{aligned}|||f(A)+f(B)|||le \frac{f(M)}{M}|||A+B||||le ||f(A+B)||| \end{aligned}$$holds for all A and B with \(M\le A+B\)。此外,我们还应用我们的结果探讨了一些相关的不等式。作为应用,我们提出了闵科夫斯基行列式不等式的一般化。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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