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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:

$\displaystyle ({{\text{I}}_1})\quad \vert(Tx,y){\vert^2} \leq (\vert T{\vert^{2\alpha }}x,x)(\vert{T^*}{\vert^{2(1 - \alpha )}}y,y)$

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 $ \vert T{\vert^{2\alpha }}x$ and $ {T^*}y$ are linearly dependent iff $ Tx$ and $ \vert{T^*}{\vert^{2(1 - \alpha )}}y$ are linearly dependent. $ ({{\text{I}}_1})$ is equivalent to

$\displaystyle ({{\text{I}}_2})\quad \vert(Tx,y)\vert \leq \vert\vert\;\vert T{\vert^\alpha }x\vert\vert\;\vert\vert\;\vert{T^*}{\vert^{1 - \alpha }}y\vert\vert,$

, so one might believe that the equality in $ ({{\text{I}}_1})$ or $ ({{\text{I}}_2})$ would hold iff $ \vert T{\vert^{2\alpha }}x$ and $ \vert{T^*}{\vert^{2(1 - \alpha )}}y$ are linearly dependent or iff $ \vert T{\vert^\alpha }x$ and $ \vert{T^*}{\vert^{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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Additional Information

DOI: https://doi.org/10.1090/S0002-9939-1986-0846001-3
Keywords: Heinz inequality, polar decomposition
Article copyright: © Copyright 1986 American Mathematical Society