1
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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 $B(H)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.