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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.