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trace inequality

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Subject: Algebra
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Posted by: Miwa Lin
on: October 11th, 2008

Let $ A,B $ be positive semidefinite, by Jensen's inequality, it is easy to see $ [tr(A^s+B^s)]^{\frac{1}{s}}\leq [tr(A^r+B^r)]^{\frac{1}{r}} $, whenever $ s>r>0 $.

What about the $ tr(A^s+B^s)^{\frac{1}{s}}\leq tr(A^r+B^r)^{\frac{1}{r}} $, is it still valid?


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clarification

On August 28th, 2010 ojs says:

I suppose that A and B are square hermitian matrixes and that s and r are real numbers. But what are the brackets supposed to represent? Sorry for my ignorance if it is obvious.

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It is open.

On October 12th, 2008 Miwa Lin says:

It is open.

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this inequality is not true.

On December 19th, 2010 Anonymous says:

this inequality is not true. It was denied some years ago. See X.Z Zhan's

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