Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. Let $M\in \mathcal{B}(F)^+$ (i.e. $M^\*=M$ and $\langle Mx\;, \;x\rangle\geq 0$ for all $x\in F$).


I want to show that $\mathcal{B}^M(F)\subseteq \mathcal{B}^{M^{1/2}}(F)$, where
$$\mathcal{B}^M(F)=\left\{S\in \mathcal{B}(F):\,\,\,\text{Im}(S^{\*}M)\subseteq \text{Im}(M)\right\},$$
$$\mathcal{B}^{M^{1/2}}(F)=\left\{S\in \mathcal{B}(F):\,\,\,\text{Im}(S^{\*}M^{1/2})\subseteq \text{Im}(M^{1/2})\right\}.$$
Thank you



Showing the following inclusion between two subalgebras of $\mathcal{B}(F)$
The following seems like overkill to me; I'd like to see a solution with less machinery. So I just give a sketch.


* As $\newcommand{\im}{\operatorname{Im}} \im(M)^\perp = \ker(M)$ we may reduce to the case when $M$ is injective and has dense range, by compressing to $\im(M)$
* Notice we can work with $S^\*$ instead of $S$.
* So the aim is to show that if $\im(SM) \subseteq \im(M)$ then also $\im(SM^{1/2}) \subseteq \im(M^{1/2})$.
* As usual, let $M^{-1}$ be the densely defined operator with $D(M^{-1}) = \im(M)$ and $M^{-1}(M\xi) = \xi$.
* Claim: $\im(SM) \subseteq \im(M)$ if and only if $M^{-1}SM$ is everywhere defined, and bounded (closed graph theorem).


Now consider the strongly continuous one-parameter semigroup $\alpha : t\mapsto M^{it} S M^{-it}$. We analytically extend this, and it turns out that $S$ is in the domain of the $\alpha\_i$ if and only if $M^{-1}SM$ is densely defined and bounded, equivalently, $\im(SM) \subseteq \im(M)$. As the domain of $\alpha\_i$ is contained in the domain of $\alpha\_{i/2}$ this appears to give the claim you want.


Reference: Ioana Ciorănescu and László Zsidó, "Analytic generators for one-parameter groups" see <https://projecteuclid.org/euclid.tmj/1178240775>
Especially section 6 at the end.


