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Proceedings of the American Mathematical Society

ISSN 1088-6826(online) ISSN 0002-9939(print)

 
 

 

A simplified proof of Heinz inequality and scrutiny of its equality


Author: Takayuki Furuta
Journal: Proc. Amer. Math. Soc. 97 (1986), 751-753
MSC: Primary 47A30
DOI: https://doi.org/10.1090/S0002-9939-1986-0846001-3
MathSciNet review: 846001
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Abstract: An operator means a bounded linear operator on a Hilbert space $H$. We give a simplified proof of the following inequality: \[ ({{\text {I}}_1})\quad |(Tx,y){|^2} \leq (|T{|^{2\alpha }}x,x)(|{T^*}{|^{2(1 - \alpha )}}y,y)\] for any operator $T$ and for any $x, y \in H$ and for any real number $\alpha$ with $0 \leq \alpha \leq 1$. In case $0 < \alpha < 1$, the equality in $({{\text {I}}_1})$ holds iff $|T{|^{2\alpha }}x$ and ${T^*}y$ are linearly dependent iff $Tx$ and $|{T^*}{|^{2(1 - \alpha )}}y$ are linearly dependent. $({{\text {I}}_1})$ is equivalent to \[ ({{\text {I}}_2})\quad |(Tx,y)| \leq ||\;|T{|^\alpha }x||\;||\;|{T^*}{|^{1 - \alpha }}y||,\], so one might believe that the equality in $({{\text {I}}_1})$ or $({{\text {I}}_2})$ would hold iff $|T{|^{2\alpha }}x$ and $|{T^*}{|^{2(1 - \alpha )}}y$ are linearly dependent or iff $|T{|^\alpha }x$ and $|{T^*}{|^{1 - \alpha }}y$ are linearly dependent, but we can give counterexamples to these mistakes. By this fact, the form of $({{\text {I}}_1})$ is more convenient than $({{\text {I}}_2})$ in order to remind us of the case when the equality in $({{\text {I}}_1})$ or $({{\text {I}}_2})$ holds.


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Keywords: Heinz inequality, polar decomposition
Article copyright: © Copyright 1986 American Mathematical Society