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<zbml>
  <query>an:05702397</query>
  <answers from="1" to="1" total="1">
  <rec>
    <an>Zbl 1186.47012</an>
    <au>Fujii, Masatoshi; Zuo, Hongliang</au>
    <ti>Matrix order in Bohr inequality for operators.</ti>
    <la>EN</la>
    <so>Banach J. Math. Anal. 4, No. 1, 21-27, electronic only (2010).</so>
    <is>ISSN 1735-8787/e</is>
    <py>2010</py>
    <dt>J</dt>
    <cc>*47A63 47B15</cc>
    <ut>Bohr inequality for operators; matrix order; parallelogram law for operators; absolute value of operators</ut>
    <ab>Summary: The classical Bohr inequality says that $|a+b|^2\le p|a|2+q|b|^2$ for all scalars $a,b$ and $p,q&gt;0$ with $\frac1p+ \frac1q=1$. The equality holds if and only if $(p-1)a=b$. Several authors discussed operator versions of the Bohr inequality. In this paper, we give a unified proof to operator generalizations of the Bohr inequality. One viewpoint of ours is a matrix inequality, and the other is a generalized parallelogram law for the absolute value of operators, i.e., for operators $A$ and $B$ on a Hilbert space and $t\ne 0$,? $$|A-B|^2+ \tfrac 1t|tA+B|^2= (1+t)|A|^2+ \big(1+\tfrac1t\big) |B|^2.$$</ab>
  </rec>
  </answers>
</zbml>

