How to prove, that the ordering on positive bounded operators agrees with ordering of their ranges?

Question

Hypothesis: Assume, that $A$ and $B$ are positive bounded operators (on some Hilbert space $H$) and $A\geq B \geq 0$. Then ${\rm range}(A) \supset {\rm range}(B)$.

The textbook "$C^*$-algebras by example" seems to be using this hypothesis.

Is the hypothesis true? How to prove it?

Answer 1

There is an easier way to prove the (weaker) result of the other answer (by Vahid Shirbisheh). Recall that every operator $T \in B(H)$ satisfies $$ \overline{\text{range}(T)} = \ker(T^*)^\perp. $$ In particular, for self-adjoint operators such as $A$ and $B$ we have \begin{align*} \overline{\text{range}(A)} = \ker(A)^\perp,\\[1ex] \overline{\text{range}(B)} = \ker(B)^\perp. \end{align*} We assume that $0 \leq B \leq A$ holds. Equivalently: for all $x\in H$ we have $0 \leq \langle Bx,x\rangle \leq \langle Ax,x\rangle$. Thus, for $x\in \ker(A)$ we have $$ \lVert B^{1/2}x\rVert^2 = \langle B^{1/2}x,B^{1/2}x\rangle = \langle Bx,x\rangle \leq \langle Ax,x\rangle = 0, $$ hence $Bx = B^{1/2}B^{1/2}x = B^{1/2}0 = 0$. It follows that $\ker(A) \subseteq \ker(B)$ holds. Taking orthogonal complements, we find $$ \overline{\text{range}(A)} \supseteq \overline{\text{range}(B)}.\tag*{$\blacksquare$} $$

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Note that the original question remains unanswered: it is still unclear whether $\text{range}(A) \supseteq \text{range}(B)$ holds, without taking closures. As a matter of fact, the answer to this question is no.

Counterexample. Let $H$ be the sequence space $\ell^2(\mathbb{N}^+)$ and let $T \in B(H)$ be an operator that is positive, injective and compact (for instance, $T$ could be the diagonal operator $e_n \mapsto \frac{1}{n}e_n$). Then $T$ is not surjective, so we may choose some unit vector $x_0\in H \setminus\text{range}(T)$.

Let $P$ be the orthogonal projection onto $\text{span}(x_0)$. Now we have $0 \leq P \leq P + T$ and $x_0\in\text{range}(P)$. Suppose, for the sake of contradiction, that $x_0\in\text{range}(P + T)$ holds. Then we may choose some $y\in H\setminus\{0\}$ such that $(P + T)y = x_0$. But now we have $Ty = x_0 - Py \in \text{span}(x_0)$. Since $T$ is injective, we also have $Ty \neq 0$, but this contradicts our assumption that $x_0 \notin \text{range}(T)$. We conclude that $\text{range}(P) \not\subseteq \text{range}(P + T)$; we only have $$ \text{range}(P) \subseteq \overline{\text{range}(P)} \subseteq \overline{\text{range}(P + T)}.\tag*{$\boxtimes$} $$

Answer 2

If $A$ and $B$ are projections, then it is easy to prove the statement. So, we reduce the problem to this case. Of course, I can only show that the closure of the image of $B$ is a subset of the closure of the image of $A$.

Let $P_A$ and $P_B$ be the support projections associated with $A$ and $B$, that is projections on the closure of the range of $A$ and $B$, respectively. Since $A$ and $B$ are positive we have $A^\alpha\to P_A$ and $B^\alpha \to P_B$ strongly as $\alpha\to 0$, see page 21 of Blackadar's book. Combining the above limits and the inequality $B^\alpha\leq A^\alpha$ for all $0<\alpha<1$ and the fact that strong convergence preserves the inequality between positive elements, we get $P_B\leq P_A$. Therefore $\overline{Im(B)}=Im(P_B)\subseteq Im(P_A)= \overline{Im(A)}$.