Let $\mathcal{H}$ be a separable Hilbert space and let $TC( \mathcal{H})$, $HS(\mathcal{H})$ be the space of trace-class operators and Hilbert-Schmidt operators on $\mathcal{H}$. Recall that these space are Banach spaces and that $HS(\mathcal{H})$ is a even a Hilbert space. For a Banach space $X$ let $B(X)$ be the space of bounded operators on $X$.


For $a \in B(TC( \mathcal{H}))$ since $TC( \mathcal{H})$ is dense in $HS(\mathcal{H})$ there is at most bounded linear extension, but it may not be bounded.
For example one may consider $a(x) = Tr(x)y $ where $y \neq 0$ is a trace-class operator.


Now, consider the set


\begin{align\*}
\mathcal{A} = \left \{ a \in B(TC( \mathcal{H})) \mid \bar a \in B(HS(\mathcal{H})) \right \} 
\end{align\*}


what can I say about this space?


It is a vector space and by the example it is not all of $ B(TC( \mathcal{H}))$.


Can I for example define an involution by taking
$\tilde a = \bar a^\*\mid\_{TC}$?



Which operators on the trace-class operators extend to operators on Hilbert-Schmidt operators?
No, this space is not closed under involution. Choose $A \in TC$ and $B \in HS\setminus TC$ and consider the rank one operator $T \mapsto \langle T, B\rangle A$ on $HS$. This restricts to a bounded operator on $TC$ but its adjoint doesn't.


