Let $1<a,b$ be parameters and $A$ be the linear operator on $L^2(\mathbb R;\mathbb C)$ defined by its kernel: $$K_A(x;y) = \int_{-1}^1 \exp\big(-|x-au|^2\big)\exp\big(-|y-bu|^2\big) \, du \,.$$ I wish to find an upper bound of the trace norm of the operator $A$, of the form (say): $$\|A\|_{\text{Tr}} \leq C \frac{1}{\sqrt{ab}}\,\,\ln(\max\{a,b\})\,,$$ for some constant $C>0$ independent of $a$ and $b$.
Would someone have an idea how to do that? (Or whether it is reasonable or not?)
What I did until now
- I got the upper bound $\|A\|_{\text{Tr}} \leq C$ for some constant $C>0$ independent of $a$ and $b$, since for every fixed $u\in\mathbb R$, $$\exp\Big(-\big|x-au\big|^2-\big|y-bu\big|^2\Big)$$ is the kernel of a rank one operator whose trace norm is independent of $a$, $b$ and $u$.
- I computed the kernel of $A^*A$ and tried unsuccessfully to compute its square root.