A\(\geq B\geq 0\) iff \((B^ rA^ pB^ r)^{1/q}\geq B^{(p+2r)/q}\) for r\(\geq 0\), p\(\geq 0\), q\(\geq 1\) with \((1+2r)q\geq p+2r\)
A capital letter means a bounded linear operator on a Hilbert space. An operator T is said to be positive if (Tx,x)\(\geq 0\) for every x in a Hilbert space H. It is well known that \(A\geq B\geq 0\) does not always ensure \(A^ 2\geq B^ 2\) in general. We investigated ''order preserving operator inequalities'', that is, we consider two operator functions f and g depending on A and B such that \(f(X)=(B^ rA^ pB^ r)^{1/q}\) and \(g(Y)=(A^ rB^ pA^ r)^{1/q}\), then \(f(A^ p)\geq f(B^ p)\) and \(g(A^ p)\geq g(B^ p)\) hold under some suitable conditions as follows. Theorem. If \(A\geq B\geq 0\), then for each \(r\geq 0\) (i) \((B^ rA^ pB^ r)^{1/q}\geq B^{(p+2r)/q}\), (ii) \(A^{(p+2)/q}\geq (A^ rB^ pA^ r)^{1/q}\) hold for each p and q such that \(p\geq 0,q\geq 1\) and \((1+2r)q\geq p+2r.\) Corollary. If \(A\geq B\geq 0\), then for each \(r\geq 0\) (i) \((B^ rA^ pB^ r)^{1/p}\geq B^{(p+2r)/p}\), (ii) \(A^{(p+2r)/p}\geq (A^ rB^ pA^ r)^{1/p}\) hold for each \(p\geq 1.\) By Theorem, we have \((BA^ 2B)^{3/4}\geq B^ 3\) and \(A^ 3\geq (AB^ 2A)^{3/4}\) for \(A\geq B\geq 0\), this fact easily implies \((BA^ 2B)^{1/2}\geq B^ 2\) and \(A^ 2\geq (AB^ 2A)^{1/2}\) for \(A\geq B\geq 0\). The latter fact is just an affirmative answer to a conjecture posed by Chan and Kwong.
scientific article; zbMATH DE number 4004981