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Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. Let $M\in \mathcal{B}(F)^+$ and set $$\mathcal{B}_{M}(F) =\left\{S \in \mathcal{B}(F):\quad \exists \,c > 0;\;\langle MSy\;, \;Sy\rangle \leq c \langle My\;,\;y\rangle,\;\forall\, y \in F \right\}.$$

Consider $M = {\rm diag}(0, 1, 0, \frac{1}{2!}, 0, \frac{1}{3!}, \dots) \in \mathcal B(F)^+$ and $ S = \left[\begin{matrix}0 & 1 \\ &0 & 1 \\ &&\ddots&\ddots\end{matrix}\right]. $

I want to show that $S\notin \mathcal{B}_{M}(F)$.

Thank you!

$M = {\rm diag}(0, 1, 0, \frac{1}{2!}, 0, \frac{1}{3!}, \dots) \in \mathcal B(F)^+$

Schüler
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1 Answers1

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Let $e_n$ be the vector in $F$ whose $n$'th coordinate is $1$ and all the rest are zero. Then, with $y=e_{2n+1}$, we have $Sy=e_{2n}$, and so $MSy=\frac{e_{2n}}{n!}$, hence $\langle MSy,Sy\rangle=\frac{1}{n!}$, whereas $My=0$. Therefore $S$ does not belong to $B_M(F)$.

uniquesolution
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