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 $\mathcal{B}^M(F)$ is a subalgebra of $\mathcal{B}(F)$, where $$\mathcal{B}^M(F)=\left\{S\in \mathcal{B}(F):\,\,\,\text{Im}(S^{*}M)\subseteq \text{Im}(M)\right\}.$$
Thank you.
There is no need for $M$ to be positive.
For any $x\in F$, $S,T\in B^M(F)$, $\alpha\in\mathbb C$, $$ (\alpha S+T)^*Mx=\bar\alpha S^*Mx+T^*Mx\in MF $$ since $MF$ is a subspace.
And, sincd $S^*Mx=My$ for some $y$ $$ (ST)^*Mx=T^*(S^*Mx)=T^*My\in MF. $$
So $B^M(F)$ is a subalgebra. It is not a $*$-subalgebra in general, even if $M$ is positive: if $M=\begin{bmatrix} 1&0\\0&0\end{bmatrix}$, $S=\begin{bmatrix} 1&0\\1&0\end{bmatrix}$, then $S^*MF\subset MF$ (in fact, $S^*M=M$), but $SMF\not\subset MF$.