Consider the function $f \in C^\infty_c(\mathbb{R}^d)$ and the operator $T_f:L^2(\mathbb{R}^d)\rightarrow L^2(\mathbb{R}^d)$ given by $g\mapsto fg$. My question: Is it true that \begin{align} \|T_f\|_\text{tr}= \operatorname{tr}(T_f^\ast T_f)^\frac{1}{2} = \int_{\mathbb{R}^d} |f(x)|\ dx? \end{align} My gut is telling me that it is true, but I couldn't find the reference I need.
If it is not true then, maybe, we have \begin{align} \|T_f\|_\text{tr}\le C\int_{\mathbb{R}^d} |f(x)|\ dx. \end{align}
First a comment: the trace norm would be $$ \|T_f\|_1=\operatorname{Tr}((T_f^*T_f)^{1/2})=\operatorname{Tr}(T_{|f|}). $$ The norm you wrote is the Hilbert-Schmidt norm, $$ \|T_f\|_2=\operatorname{Tr}(T_f^*T_f)^{1/2}=\operatorname{Tr}(T_{|f|^2})^{1/2}. $$ In any case, both will be infinity as long as $f\ne0$. A multiplication operator is never compact when the measure is diffuse.
To see it, let $V\subset\mathbb R^d$ be open with $|f|>\delta>0$ on $V$ (exists by continuity and $f\ne0$). Fix $m\in\mathbb N$. Partition $V$ in $m$ measurable subsets $V_1,\ldots,V_m$, and let $$ e_k=\frac1{|V_k|^{1/2}}\,1_{V_k}. $$ Then $e_1,\ldots,e_m$ are orthonormal. And we have $$ \operatorname{Tr}(T_{|f|})\geq\sum_k\langle T_{|f|}e_k,e_k\rangle =\sum_k\frac1{|V_k|}\int_{V_k}|f|\geq\sum_k\delta=m\delta. $$ As we can do this for any $m$, $\operatorname{Tr}(T_{|f|})=\infty$. With the exact same idea, $\operatorname{Tr}(T_{|f|^2})=\infty$.