Can the positive operator stay monotonicity？

Question

If $A,B$ are two self-adjoint operators on Hilbert space $\mathscr{H}$, and assume $A\ge B \ge0$, which means that $$ ((A-B)f,f)\ge 0, \forall f \in D(A) \cap D(B), $$ and $$ \quad(Af_1,f_1) \ge0, (Bf_2,f_2) \ge0, \forall f_1 \in D(A), \;f_2 \in D(B), $$ then can we conclude that $$ A^s \ge B^s, \;\;\forall s>0? $$ where we can use spectral decomposition theorem to define $A^s, B^s$. The result is easy for $\dim \mathscr{H}< \infty$, but when it comes to finite case, the continuous spectrum may occur, and then I am a little confused about how to compare them?

Answer 1

Here is a counter-example for $s=2$. We work in $H=\mathbb{C}^2$, so $B(H)\cong M_2(\mathbb{C})$.

Simply take $a=\begin{pmatrix}1&-1\\-1&1\end{pmatrix}$ and $b=\begin{pmatrix}7&2\\2&1\end{pmatrix}$. These are both positive matrices and note that $0\leq a\leq a+b$, where $a+b=\begin{pmatrix}8&1\\ 1&2\end{pmatrix}$. Note that

$$a^2=\begin{pmatrix}2&-2\\ -2&2\end{pmatrix} $$ $$(a+b)^2=\begin{pmatrix}65&10\\10&5\end{pmatrix} $$ However, $(a+b)^2-a^2=\begin{pmatrix}63&12\\12&3 \end{pmatrix}$ which is not positive (compute the eigenvalues and you can see it). Note that neither $a^2-(a+b)^2$ is positive! In particular, $(a+B)^2-a^2$ has two distinct eigenvalues, one positive and one negative.