In most literature, $\mathcal{J}_1$ is the set of operators that are trace class, i.e. $\mathcal{J}_1=\{\text{Operator }T\,:\,||T||_1<\infty\}$, where $||\cdot||_1$ denotes the Schatten-1 norm.
Let $A$ be some integral-operator acting on $\mathcal{H}$ as in the title. That is to say, for $f\in \mathcal{H}$, \begin{align} Af(x)=\int_{\mathbb{R}^2}K(x,y)f(y)\,\mathrm{d}y\,, \end{align} for its kernel $K$. In the affirming case of $A$ being trace-class, a result in Trace ideals and their applications by Barry Simon (I think this result is seen in a lot of places, really) states that \begin{align} \text{Tr }A = \int_{\mathbb{R}^2}K(x,x)\,\mathrm{d}x\,.\quad (1) \end{align} My question is; are there any tools out there to actually show whether such an integral operator is trace class, before moving on to calculate the trace? Ironically, I believe that in my particular case I have bounds on $K$ to show that $(1)$ is finite. However, because $||\cdot||$ is usually formulated with the operator $|A|$, I find it difficult to show that an integral operator is trace class merely from the definition.
To compare it to another setting, let $\mathcal{H}'=\ell^2(\mathbb{Z}^2)$, the lattice with orthonormal basis consisting of $\{\left|x\right>\}$ where \begin{align} \left|x\right>:\ell^2(\mathbb{Z}^2)&\rightarrow \mathbb{C}\\ y&\mapsto \delta_{x,y}\,. \end{align} Here, we can talk about an operator $T$ having matrix elements $T_{x,y}=\langle x\,,Ty\rangle $. A result in https://arxiv.org/pdf/cond-mat/9603116.pdf (Lemma 1, it is stated for $p=3$ but the proof also works for $p=1$) says that \begin{align} ||T||_1\leq \sum_{b\in \mathbb{Z}^2}\sum_{x\in \mathbb{Z}^2} |T_{x+b,x}|\,. \end{align} So to prove $T$ is trace class in this situation, all I need to do is show that the r.h.s in the above is finite. Are there similar tools in the continuum case?