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Define $$\operatorname{Ref}\mathcal{S}=\{T\in B(\mathcal{H}):Th\in[\mathcal{S}h], \forall h \in \mathcal{H}\},$$where $\mathcal{H}$ is a Hilbert space and $\mathcal{S}$ is a linear manifold of $B(\mathcal{H})$.

A proposition of Conway's book A Course in Operator Theory says that $\operatorname{Ref}\mathcal{S^\ast}=(\operatorname{Ref}\mathcal{S})^\ast$ and the proof is left as an easy exercise. It is not easy for me, thanks to the one who can tell me a proof or give me a hint.

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  • $\begingroup$ What is $[\mathcal{S}h]$? $\endgroup$ – Owen Sizemore Jul 9 '13 at 2:50
  • $\begingroup$ Notation is a little funny. The two $*$'s in the equation don't mean the same thing, adjoint of operators on the left and Hilbert space duals on the right. You should explain that. $\endgroup$ – Michael Jul 9 '13 at 3:20
  • $\begingroup$ @ Owen Sizemore: $[\mathcal{S}h]$ is the closure of $span\{Sh:S\in\mathcal{S}\}$. $\endgroup$ – Zhonghua Wang Jul 9 '13 at 3:58
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    $\begingroup$ @ Michael: The two *'s are both adjoints of operators. $\endgroup$ – Zhonghua Wang Jul 9 '13 at 4:00
  • $\begingroup$ Where is this in the book? $\endgroup$ – Jonas Meyer Jul 9 '13 at 6:14
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Suppose $T \in Ref(\mathcal{S})$, we want to show that $T^{\ast} \in Ref(\mathcal{S}^{\ast})$. ie. For any $h \in H$, we want to show that $T^{\ast}h \in [\mathcal{S}^{\ast}h]$. Since $[\mathcal{S}^{\ast}h]$ is closed, it suffices to show that for any linear functional $\varphi$ on $H$, $$ \varphi([\mathcal{S}^{\ast}h]) = 0 \Rightarrow \varphi(T^{\ast}h) = 0 $$ By Riesz Representation, it suffices to show that, for any $y \in H$, $$ \langle S^{\ast}h, y\rangle = 0 \quad\forall S\in \mathcal{S} \Rightarrow \langle T^{\ast}h,y\rangle = 0 $$ $$ \Leftrightarrow \langle h, Sy\rangle = 0 \quad\forall S\in \mathcal{S} \Rightarrow \langle h, Ty\rangle = 0 $$ But for any $y \in H$, $Ty \in [\mathcal{S}y]$, and so this is true. Hence, $T^{\ast} \in Ref(\mathcal{S}^{\ast})$ and so $$ Ref(\mathcal{S})^{\ast} \subset Ref(\mathcal{S}^{\ast}) $$ The argument is similar for the other containment.

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  • $\begingroup$ Indeed, $\operatorname{Ref}(\mathcal{S})$ consists precisely of the $T \in B(\mathcal{H})$ such that, for all $h,y \in \mathcal{H}$, one has $$\big[\langle Sh,y \rangle = 0 \ \ \forall S \in \mathcal{S} \big] \ \ \Rightarrow \ \ \langle Th,y \rangle =0.$$ $\endgroup$ – Mike F Oct 8 '13 at 7:51

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