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.

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 $\begin{pmatrix} 2 \\ 0 \end{pmatrix}\notin \mathcal{R}(M)$.

• Thank you for your answer. If we replace $\mathbb{R}^2$ by $\mathbb{C}^2$. Is the example remains true? – Student Nov 5 '17 at 11:44
• @Student I edited to adress the complex case. I didn't read your question careful enough in the beginning. – Severin Schraven Nov 5 '17 at 13:54
• Thank you very much, but I don't understand why M is positive with respect to the complex inner product? – Student Nov 5 '17 at 14:15
• @Student Which step is not clear? – Severin Schraven Nov 5 '17 at 15:11
• Sorry all is clear. Thank you – Student Nov 5 '17 at 15:20