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Norm of Schur multiplication for Schatten norm


Author: K. Okubo
Journal: Proc. Amer. Math. Soc. 111 (1991), 925-930
MSC: Primary 15A60; Secondary 15A04, 47A30
DOI: https://doi.org/10.1090/S0002-9939-1991-1041011-X
MathSciNet review: 1041011
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Abstract: Let $ {\mathbb{M}_n}$ denote the algebra of all $ n \times n$ complex matrices and $ \vert\vert \cdot \vert\vert$ be the Schatten's $ p$-norm on $ {\mathbb{M}_n}$. For each $ A \in {\mathbb{M}_n}$, a linear operator $ {S_A}$ on $ {\mathbb{M}_n}$ is defined by $ {S_A}\left( X \right): = A \circ X$ for all $ X \in {\mathbb{M}_n}$, where $ \circ $ denotes the Schur product and $ \vert\vert{S_A}\vert{\vert _{p,q}}$ is defined as the operator norm from $ ({\mathbb{M}_n},\vert\vert \cdot \vert{\vert _p})$ to $ ({\mathbb{M}_n},\vert\vert \cdot \vert{\vert _q})$ for $ p,q \geq 1$.

Given an $ A \in {\mathbb{M}_n}$, suppose $ 0 < \lambda < 1$, and $ p,{p_1},{p_2},q,{q_1}$, and $ {q_2}$ are not smaller than 1 , and

$\displaystyle \frac{1} {p} = \frac{\lambda } {{{p_1}}} + \frac{{1 - \lambda }} ... ...quad \frac{1} {q} = \frac{\lambda } {{{q_1}}} + \frac{{1 - \lambda }} {{{q_2}}}$

are satisfied. Then we will show that $ \vert\vert{S_A}\vert{\vert _{p,q}} \leq \vert\vert{S_A}\vert\vert _{{p_1},{q_1}}^\lambda \cdot \vert\vert{S_A}\vert\vert _{{p_2},{q_2}}^{1 - \lambda }$.

References [Enhancements On Off] (What's this?)

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Additional Information

DOI: https://doi.org/10.1090/S0002-9939-1991-1041011-X
Keywords: Schur multiplication, Schatten norm
Article copyright: © Copyright 1991 American Mathematical Society

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