Let $\mathcal{H}$ be the Hilbert space $L^2(\mathbb{R})$. View the Fourier transform as a unitary operator $\mathcal{F} \in B(\mathcal{H})$.
For each function $f \in C_0(\mathbb{R})$, let $T(f) \in B(\mathcal{H})$ be the multiplication operator given as $(T(f) \xi)(t) = f(t) \xi(t)$ for all $\xi \in \mathcal{H}$, $t \in \mathbb{R}$.
- On the other hand, we can conjugate by the Fourier transform to get another representation of $C_0(\mathbb{R})$ on $\mathcal{H}$. Define $S(f) = \mathcal{F} T(f) \mathcal{F}^{-1}$ for each $f \in C_0(\mathbb{R})$. To describe this map another way, note that if $f$ is actually the Fourier transform of $h \in L^1(\mathbb{R})$, then $S(f)$ is convolution with $h$ i.e. $(S(f) \xi)(t) = \int h(s) \xi(t-s) \ dt$ for all $\xi \in \mathcal{H}$, $t \in \mathbb{R}$.
For each pair of functinos $f,g \in C_0(\mathbb{R})$, let us define $$ K(f,g) := T(f) S(g) = T(f) \ \mathcal{F} \ T(g) \ \mathcal{F}^{-1}.$$ Now I've been able to show the following things:
- For all $f,g \in C_0(\mathbb{R})$, the operator $K(f,g)$ is compact.
- If $f,g \in C_0(\mathbb{R}) \cap L^2(\mathbb{R})$, then $K(f,g)$ is a Hilbert-Schmidt operator. Moreover, if $f_1,f_2, g_1,g_2 \in C_0(\mathbb{R})$, then $$\operatorname{tr}\bigg(K(f_1,g_1)^* K(f_2,g_2) \bigg) = \frac{1}{2 \pi} \left( \int \overline{f_1(t)} f_2(t) \ dt \right) \left( \int \overline{ g_1(t)} g_2(t) \ dt \right)$$
I strongly suspect that the following is also true, but I have no idea how to prove it!?
If $f, g \in C_0(\mathbb{R}) \cap L^1(\mathbb{R})$, then $K(f,g)$ is trace-class. If the latter is true, then it must also be true that $$\operatorname{tr}(K(f,g)) = \frac{1}{2 \pi } \left( \int f(t) \ dt \right) \left( \int g(t) \ dt \right)$$ I also have an inkling that the bound $$ \operatorname{tr}\left( | K(f,g) | \right) \leq \|f\|_1 \|g\|_1$$ may be valid where, for an operator $A \in B(\mathcal{H})$, one writes $|T|$ for the operator $\sqrt{A^*A} \in B(\mathcal{H})$. Knowing this would also answer my question.