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 $\mathcal{B}_1(F)$ is not a subalgebra of $\mathcal{B}(F)$, where $$\mathcal{B}_1(F)=\left\{S\in \mathcal{B}(F):\,\,\exists c>0 ;\;\langle MSy\;, \;Sy\rangle \leq c \langle My\;,\;y\rangle,\;\forall y \in \overline{\text{Im}(M)}\right\}.$$

Thank you.


1 Answer 1


Let $M = {\rm diag}(0, 1, 0, \frac{1}{2!}, 0, \frac{1}{3!}, \dots) \in \mathcal B(F)^+$ and $S\in \mathcal B(F)$ be the backward unilateral shift, $$ S = \left[\begin{matrix}0 & 1 \\ &0 & 1 \\ &&\ddots&\ddots\end{matrix}\right]. $$ It is easy to calculate that $MSM = 0$ and so $MSy = 0$ for all $y\in \overline{{\rm Im}(M)}$. Hence, $S\in \mathcal B_1(F)$. However, for every $n\geq 2$ we have \begin{align*} \langle MS^2Me_{2n}, S^2Me_{2n}\rangle & = \langle MS^2\frac{1}{n!}e_{2n}, S^2\frac{1}{n!}e_{2n}\rangle \\ & = \frac{1}{(n!)^2}\langle M e_{2(n-1)}, e_{2(n-1)}\rangle \\ & = \frac{1}{(n!)^2(n-1)!} \end{align*} but \begin{align*} \langle MMe_{2n}, Me_{2n}\rangle = \frac{1}{(n!)^3}. \end{align*} Thus, $$ \langle MS^2Me_{2n}, S^2Me_{2n}\rangle = n\langle MMe_{2n}, Me_{2n}\rangle $$ and so $S^2 \notin \mathcal B_1(F)$. Therefore, $\mathcal B_1(F)$ is not a subalgebra of $\mathcal B(F)$.


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