Adjoint of the attached space in operator algebras

Question

J. Conway considers in his book "A course in operator theory" a linear subspace $\mathcal S \subset B(\mathcal H)$ of bounded operators acting on a Hilbert space. He defines the attached space as $$ \mathrm{Ref} \ \mathcal S = \{ T \in B(\mathcal H) \ | \ \forall h \in \mathcal H \ : \ Th \in \mathrm{cl} (\mathcal S h) \}, $$ where $\mathrm{cl}$ denotes closure of a set and $\mathcal S h$ is the set of all vectors of the form $Ah$ with $A \in \mathcal S$. He leaves as a simple exercise to prove that $\mathrm{Ref} \ (\mathcal S^*)=(\mathrm{Ref} \ \mathcal S)^*$ - that is space attached to the set of adjoint operators consists precisely of adjoints of elements in the attached space. I have no idea how to proceed here. It seems that it is necessary to somehow relate ranges of an operator and its adjoint, but such relation certainly doesn't exist - instead range is connected with kernel. Explanation of this equality or hints with be appreciated

Answer 1

You can rewrite the definition in the following way: for any $h$, whenever a vector $v\in H$ is orthogonal to the subspace $\mathcal{S}h$, then it is also orthogonal to $Th$.

Let me show now that $(\textrm{Ref}\mathcal{S})^{\ast} \subset \textrm{Ref}\mathcal{S}^{\ast}$. Take $T^{\ast}$ for some $T \in \textrm{Ref}\mathcal{S}$. Fix $h \in H$ and consider a vector $v$ orthogonal to $\mathcal{S}^{\ast}h$. It means that for any $S\in\mathcal{S}$ we have $\langle v, S^{\ast} h \rangle = 0 = \langle Sv, h\rangle$. It follows that $h$ is orthogonal to $\mathcal{S}v$, hence it is also orthogonal to $Tv$, i.e. $\langle Tv,h\rangle = 0 = \langle v, T^{\ast} h\rangle$; we are done. The other inclusion also follows as we get from this argument that $(\textrm{Ref} \mathcal{S}^{\ast})^{\ast} \subset \textrm{Ref}(\mathcal{S}^{\ast})^{\ast} = \textrm{Ref}\mathcal{S}$.