Abstract
The classical Bohr inequality says that $|a+b|^2\leq p|a|^2+q|b|^2$ for all scalars $a, \ b$ and positive $p,q$ with $\frac 1p + \frac 1q =1.$ The equality holds if and only if $(p-1)a=b.$ Several authors discussed operator version of Bohr inequality. In this paper, we give a unified proof to operator generalizations of Bohr inequality. One viewpoint of ours is a matrix inequality, and the other is a generalized parallelogram law for absolute value of operators, i.e., for operators $A$ and $B$ on a Hilbert space and $t\neq0$, $$|A-B|^2+\frac{1}{t}|tA+B|^2=(1+t)|A|^2+(1+\frac{1}{t})|B|^2.$$
Citation
Masatoshi Fujii. Hongliang Zuo. "Matrix order in Bohr inequality for operators." Banach J. Math. Anal. 4 (1) 21 - 27, 2010. https://doi.org/10.15352/bjma/1272374669
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