Showing the following inclusion between two subalgebras of $\mathcal{B}(F)$ 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
 A: 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.
