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Just to avoid confusion, a function is called matrix monotone in an interval $[a, b]$ if $A - B \geq 0$ implies $f(A) - f(B) \geq 0$ for any Hermitian Matrices $A, B$ (we can restrict to finite dimensions) with spectrum in $[a, b]$. ($\geq 0$ means that the matrix is positive semi-definite)

I am currently interested in the function $f(t) = 1 - ( 1 - t^{1/2} )^2$ and the interval $[0, 1]$. The function is monotonically increasing and concave in this interval, so there is reason to hope that it is also operator monotone.

However, I do not know how to prove this and was wondering if there exists a general strategy to prove operator monotonicity of functions. Any ideas?

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up vote 3 down vote accepted

Operator monotonicity is rather tricky. Counterexamples will generally involve non-commuting matrices. On the other hand, you can assume one of the matrices is diagonal. Another clue is that often the extreme cases are the ones to look at. So I'll try $ B = \pmatrix{b & 0\cr 0 & 0\cr}$, $A = \pmatrix{a & c\cr c & d\cr}$, where $b = \dfrac{c^2+a^2-a}{c^2+a-1}$ and $d = 1 - \dfrac{c^2}{1-a}$ and $0 \le a \le 1 - c^2$. After some experimentation, I found that with, say, $a=4/5$ and $c=1/5$,

$$ A = \pmatrix{4/5 & 1/5\cr 1/5 & 4/5\cr},\ B = \pmatrix{3/4 & 0\cr 0 & 0\cr}$$ $$ f(A) = \pmatrix{\frac{1}{5}+\sqrt{\frac{3}{5}} & \frac{4}{5} - \sqrt{\frac35}\cr \frac{4}{5} - \sqrt{\frac35} & \frac{1}{5}+\sqrt{\frac{3}{5}}\cr},\ f(B) = \pmatrix{\sqrt{3}-\frac34 & 0\cr 0 & 0\cr}$$ and $f(A)_{11} < f(B)_{11}$. So this function is not operator monotone.

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Thanks @Robert. This is very helpful and solves my immediate problem. Still, is there a generic strategy to answer the question whether a function is operator monotone in the positive? – Marco Tomamichel Jun 15 '12 at 4:54

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