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.
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.