# On the equality Tr(Af) = Tr(fA)

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

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 $B(H)$.