trace inequality

Importance: Medium ✭✭
Author(s):
Subject: Algebra
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Recomm. for undergrads: no
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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Bibliography

% Example: %*[B] Claude Berge, Farbung von Graphen, deren samtliche bzw. deren ungerade Kreise starr sind, Wiss. Z. Martin-Luther-Univ. Halle-Wittenberg Math.-Natur. Reihe 10 (1961), 114. % %[CRS] Maria Chudnovsky, Neil Robertson, Paul Seymour, Robin Thomas: \arxiv[The strong perfect graph theorem]{math.CO/0212070}, % Ann. of Math. (2) 164 (2006), no. 1, 51--229. \MRhref{MR2233847} % % (Put an empty line between individual entries)


* indicates original appearance(s) of problem.

clarification

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.

It is open.

It is open.

this inequality is not true.

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

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