Consider the Hilbert space $H = L^2(\mathbb{R})$, and a bounded operator $A \in B(H)$ which satisfies:
$$
\forall f \in H, \quad Af \text{ is trace class and } Tr(Af) < C \| f \|\_{H},
$$
where $f$ is seen as the multiplicative operator by the function $f$. Can we deduce
$$
\forall f \in H, \quad fA \text{ is trace class},
$$
which will automatically imply that $Tr(Af) = Tr(fA)$? 


Does someone know whether such a result has been investigated, and if yes, what was the result?
Thanks



On the equality Tr(Af) = Tr(fA)
The answer to your original question is NO. Here is an counterexample: 


For simplicity, assume everything is real-valued and the Hilbert space is over $\mathbb{R}$, let $$ \varphi \in L^2(\mathbb{R}) \setminus L^\infty(\mathbb{R}) \text{ and } \varphi\chi\_{[0,1]}\in L^\infty([0,1]);$$
$$\psi \in L^2([0,1]) \cap L^\infty([0,1]).$$ 


Consider the rank one operator $A = \varphi \otimes \psi$, i.e., 
$$
A(\xi) = \varphi \langle \xi, \psi\rangle\_{L^2(\mathbb{R})}.
$$
Then $A$ verifies of course your assumption. Indeed, we have 
$$
Af = \varphi \otimes (\psi\cdot f) \text{ and } \mathrm{tr} (Af) = \int\varphi(x) \psi(x) f(x) dx \le \|\varphi\psi \|\_{L^2([0,1]} \| f\|\_{L^2([0,1])} \le C\| f\|\_H. 
$$
But now, for $fA$ to be in $B(H)$, it is necessary that for any $\xi \in H$, we have
$$
 f \in H \Longrightarrow f \cdot A(\xi) \in H,
$$
this means that $A(\xi) \in L^\infty(\mathbb{R}) \cap L^2(\mathbb{R})$. But by our choice of $\varphi$, we know that
$$
A(\psi) = \| \psi\|\_H^2\cdot \varphi \notin L^\infty(\mathbb{R}) \cap L^2(\mathbb{R}).
$$
This means that $fA$ is not even in $B(H)$.


