Positive bounded operators

Question

Let $A,B$ be positive self-adjoint bounded operators and $\lambda >0$ then I want to show that if

$$A-B \ge 0 $$ that is $\langle x,(A-B)x \rangle \ge 0$ we have that the resolvents (whose existence is clear) satisfy $$(A+\lambda I)^{-1}-(B+\lambda I)^{-1} \le 0,$$ i.e. exactly the opposite relation. Although this is intuitively clear, I got nowhere by applying the definition. I guess there is some trick to flip this inequality over to the other.

@TrialAndError I think this is something where you immediately know how to do this. — Zimkovic, Nov 18 '15 at 01:46

Martin Argerami · Accepted Answer · 2015-11-18T19:31:23.530

2

We immediately have that $B+\lambda\leq A+\lambda$ and that both are invertible. So all we need to show is that if $X\leq Y$ with $X$ and $Y$ invertible, then $Y^{-1}\leq X^{-1}$.

We start with $X\leq Y$. By conjugating with $Y^{-1/2}$ (which is also positive), we get $$ 0\leq Y^{-1}XY^{-1/2}\leq I, $$ which we can rewrite as $$ 0\leq (X^{1/2}Y^{-1/2})^*X^{1/2}Y^{-1/2}\leq I. $$ A positive operator is between $0$ and $I$ precisely when its spectrum is contained in $[0,1]$. And, for invertible $R$, the spectrum of $R^*R$ is the same as that of $RR^*$. Thus $$ 0\leq X^{1/2}Y^{-1/2}(X^{1/2}Y^{-1/2})^*\leq I. $$ This can be rewritten as $$ X^{1/2}Y^{-1}X^{1/2}\leq I. $$ Conjugating with $X^{-1/2}$ we get $$ Y^{-1}\leq X^{-1}. $$

edited Nov 18 '15 at 19:31

answered Nov 18 '15 at 03:07

Martin Argerami

205,756

The $3/2$ in $(A+\lambda I)^{-1}$ should be $3/5$. So the $5/2$ in the final matrix should be $-2/5$. And the $9/10$ in your final matrix should be $2/5-1/2=4/10-5/10=-1/10$. The final matrix is then negative. – Disintegrating By Parts Nov 18 '15 at 11:44
Yes. I took the $1$ from a calculation I made but I evaluated my formulas wrong... – Martin Argerami Nov 18 '15 at 17:15
@MartinArgerami well yes, as you see in my question I am only interested in the case $\lambda >0.$ Although your symbolic argument $X \le Y$ then $Y^{-1} \le X^{-1}$ is also what I used, would you mind giving a rigorous argument based on the definition of $X \ge Y$ if and only if $\langle x,(X-Y) x \rangle \ge 0$? – Zimkovic Nov 18 '15 at 18:11
I guess I missed that $\lambda>0$. I'll post the argument later, as right now I have to go teach. – Martin Argerami Nov 18 '15 at 18:23
I'm back, and I have edited the answer. – Martin Argerami Nov 18 '15 at 19:31
Much better answer. +1. – Disintegrating By Parts Nov 18 '15 at 20:17

Positive bounded operators

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