Showing that this inclusion between two subspaces of $\mathcal{B}(\mathcal{H})$ can be strict. Let $\mathcal{H}$ be a complex Hilbert space, with inner product $\langle\cdot\;| \;\cdot\rangle$ and the norm $\|\cdot\|$ and 
let $\mathcal{B}(\mathcal{H})$ be the algebra of all bounded linear operators from $\mathcal{H}$ to $\mathcal{H}$.
Let $M\in \mathcal{B}(\mathcal{H})^+$ (i.e. $M^*=M$ and $\langle Mx\;| \;x\rangle \geq0,\;\forall x\in \mathcal{H}$), we consider the following two subspaces of $\mathcal{B}(\mathcal{H})$:
$$\mathcal{B}_1(\mathcal{H})=\left\{T\in \mathcal{B}(\mathcal{H}):\,\,\,\mathcal{R}(T^{*}M)\subseteq \mathcal{R}(M)\right\}.$$
$$\mathcal{B}_2(\mathcal{H})=\left\{T\in \mathcal{B}(\mathcal{H}):\,\,\exists c>0 \quad \mbox{such that}\quad\langle MTx\;| \;Tx\rangle \leq c  \langle Mx\;| \;x\rangle,\;\forall x \in \overline{\mathcal{R}(M)}\right\},$$
where $\mathcal{R}(T^{*}M)$, $\mathcal{R}(M)$ are respectively the ranges of $T^{*}M$ and $M$. I see in a paper that $\mathcal{B}_1(\mathcal{H})\subsetneq \mathcal{B}_2(\mathcal{H})$. But if $M$ is injective with closed range, then $\mathcal{B}_1(\mathcal{H})=\mathcal{B}_2(\mathcal{H})=\mathcal{B}(\mathcal{H})$. 
My goal is to construct an operator $T\in \mathcal{B}_2(\mathcal{H})$ (perhas a matrix if $\mathcal{H}$ is finite dimentional) such that $T\notin \mathcal{B}_1(\mathcal{H})$.
Thank you.
 A: Take $\mathcal{H}= \mathbb{C}^2$ with the standard scalar product and choose the operators given by the following matrices
$$ M = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}, \qquad T = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}.$$
As $M$ is real and symmetric, we have $M^* =M$. Furthermore, we have
$$ \left\langle M \begin{pmatrix} x \\ y \end{pmatrix}, \begin{pmatrix} x \\ y \end{pmatrix} \right\rangle = \left\langle  \begin{pmatrix} x + y \\ x+ y \end{pmatrix}, \begin{pmatrix} x \\ y \end{pmatrix} \right\rangle = 
(x+y)(\overline{x}+\overline{y}) = \vert x+y \vert^2 \geq 0.$$
Note that $\overline{\mathcal{R}(M)}= \mathcal{R}(M) = \left\{ \begin{pmatrix} x \\ x \end{pmatrix} \ : \ x\in \mathbb{C} \right\}$ (subspaces in finite dimensions are always closed). We compute
$$ \left\langle M T\begin{pmatrix} x \\ x \end{pmatrix}, T \begin{pmatrix} x \\ x \end{pmatrix} \right\rangle 
= \left\langle M \begin{pmatrix} x \\ 0 \end{pmatrix},  \begin{pmatrix} x \\ 0 \end{pmatrix} \right\rangle 
= \left\langle \begin{pmatrix} x \\ x \end{pmatrix},  \begin{pmatrix} x \\ 0 \end{pmatrix} \right\rangle = \vert x \vert^2 \leq 4 \vert x \vert^2 
= \left\langle \begin{pmatrix} 2x \\ 2x \end{pmatrix},  \begin{pmatrix} x \\ x \end{pmatrix} \right\rangle
= \left\langle M \begin{pmatrix} x \\ x \end{pmatrix}, \begin{pmatrix} x \\ x \end{pmatrix} \right\rangle.$$
Hence, $T\in \mathcal{B}_2(\mathcal{H})$. In addition $T\notin \mathcal{B}_1(\mathcal{H})$, as (now we use that $T=T^*$ as $T$ is symmetric and real)
$$ \begin{pmatrix} 2 \\ 0 \end{pmatrix} = T\begin{pmatrix} 2 \\ 2 \end{pmatrix} = T^*\begin{pmatrix} 2 \\ 2 \end{pmatrix} = T^* M \begin{pmatrix} 1 \\ 1 \end{pmatrix} \in \mathcal{R}(T^* M),$$
but $\begin{pmatrix} 2 \\ 0 \end{pmatrix}\notin \mathcal{R}(M)$.
