Consider an integral operator on $L^2(\mathbb{R})$ with kernel $K : \mathbb{R}^2 \to \mathbb{C}$. My question is if it is possible to bound the trace-class norm of $T$ based on properties of $K$?
This problem appears to be have been studied in the case of $L^2(0,1)$ but I haven't been able to find anything in the non-compact case apart from a remark in "Kernels of trace-class operators" by Chris Brislaw saying that if the kernel function is in the Schwarz class, the operator is trace-class.
I will assume that the operator you speak of is trace-class. Now, it turns out that there is no way to bound the trace norm for some trace class operator $A$, i.e. $\|A\|_{1}$ in the same sense that we can not bound the $L_{1}(\mathbb{R})$ norm for $L_{p}(\mathbb{R})$ spaces. Schatten norms $\|A\|_{p}:= \big(Tr|A|^{p}\big)^{\frac{1}{p}}$ follow the same inclusion pattern as $L_{p}(\mathbb{R})$ spaces do. i.e $$ L_{p}(\mathbb{R}) \subset L_{q}(\mathbb{R}) $$ when $p>q$. It also turns out that Hoelder's inequality holds for these norms. Consequently
$$ \|A\|_{\infty}\leq... \leq \|A\|_{2}\leq\|A\|_{1}$$
where $\|A\|_{\infty}$ is just the operator norm. It turns out that all of the even cases are easy to compute while the odd cases are pretty much impossible without knowing the spectral decomposition of your integral's kernel. Why? Because you must compute the absolute value of an integral operator.
Now, if your integral operator $T$ is positive then $\|T\|_{1} = Tr\{T\}$ and the trace is easily computed. Also, if you can argue that your operator is a product of two Hilbert Schmidt operators (Schatten class $\|...\|_{2}$) and you know what they are then using Hoelder's should do the trick because then you would only have to take a Hilbert Schmidt norm $\|...\|_{2}$ and that is easy.