Characterization of positive operators below an operator

Question

Let $T\in B(H)$ be positive. Is there a way to characterize all positive $S$ such that $S\leq T$?

Answer 1

Suppose that $S,T\in B(H)$ with $T\geq0$. Then the following statements are equivalent:

1. there exists a contraction $R$ such that $S^{1/2}=RT^{1/2}=T^{1/2}R^*$;

2. There exists a contraction $R$ such that $RT^{1/2}=T^{1/2}R^*$ and $S=RTR^*$

3. there exists a contraction $R$ such that $S=T^{1/2}R^*RT^{1/2}$

4. $0\leq S\leq T$.

Proof. The implication 1$\implies$2 is simply $$S=S^{1/2}S^{1/2}=RT^{1/2}T^{1/2}R^*=RTR^*.$$

2$\implies$3. We have $$ S=RT^{1/2}T^{1/2}R^*=T^{1/2}R^*RT^{1/2}. $$

3$\implies$4. Since $R$ is a contraction, $$\langle R^*Rx,x\rangle=\|Rx\|^2\leq\|x\|^2=\langle x,x\rangle,$$ so $R^*R\leq I$. Then $$ S=T^{1/2}R^*RT^{1/2}\leq T^{1/2}IT^{1/2}=T. $$

4$\implies$1. Note that $\ker S^{1/2}=\ker S$, $\ker T^{1/2}=\ker T$. If $x\in \ker T^{1/2}$, then $$\tag1 \|S^{1/2}x\|^2=\langle Sx,x\rangle\leq\langle Tx,x\rangle=\|T^{1/2}x\|^2=0. $$ Thus $\ker T^{1/2}\subset \ker S^{1/2}$. Define $R$ on $T^{1/2}H$ by $$ R(T^{1/2}x)=S^{1/2}x. $$ This is well-defined, because if $T^{1/2}x=T^{1/2}y$, then $S^{1/2}x=S^{1/2}y$ by $(1)$. Also, $$ \|R(T^{1/2}x)\|^2=\|S^{1/2}x\|^2=\langle Sx,x\rangle\leq\langle Tx,x\rangle=\|T^{1/2}x\|^2, $$ so $\|R\|\leq 1$. So we may extend $R$ to $\overline{T^{1/2}H}=(\ker T^{1/2})^\perp=(\ker T)^\perp$. For $x\in\ker T$, we define $Rx=0$. Then $R\in B(H)$ is a contraction, and $RT^{1/2}=S^{1/2}$. As $S^{1/2}$ is selfadjoint, we get $RT^{1/2}=T^{1/2}R^*$.