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$\newcommand{\Tr}{\operatorname{Tr}}$Let $\mathcal{S}(\mathbb{R}^k)$ denote the $k$-dimensional Schwartz space with the usual topology, and let $\mathcal{S}'(\mathbb{R}^{k}))$ denote its strong dual (i.e. the space of tempered distributions equipped with the topology of uniform convergence on bounded sets). Let $\mathcal{S}(\mathbb{R}^k) \mathbin{\hat{\otimes}_\pi} \mathcal{S}(\mathbb{R}^k)$ denote the completed projective tensor product of $\mathcal{S}(\mathbb{R}^k)$ and $\mathcal{S}'(\mathbb{R}^k)$. Note that since both the Schwartz space and the space of tempered distributions are nuclear, the projective tensor product coincides with the injective tensor product.

If $f\in\mathcal{S}(\mathbb{R}^k)$ and $g\in\mathcal{S}'(\mathbb{R}^k)$, then we can define $$\Tr(f\otimes \bar{g}) := \overline{\langle{g, \bar{f}}\rangle}_{\mathcal{S}'-\mathcal{S}},$$ where $\langle{\cdot,\cdot}\rangle_{\mathcal{S}'-\mathcal{S}}$ denotes the duality pairing. Now if the duality pairing were a continuous map $$\mathcal{S}(\mathbb{R}^{k}) \times \mathcal{S}'(\mathbb{R}^{k}) \rightarrow \mathbb{C},$$ then by the universal property of the $\pi$-tensor product, we would obtain a unique continuous map $$\Tr: \mathcal{S}(\mathbb{R}^k) \mathbin{\hat{\otimes}_\pi} \mathcal{S}'(\mathbb{R}^k) \rightarrow \mathbb{C}$$ with the property that $\Tr(f\otimes \bar{g})$ is as above.

Unfortunately, the duality pairing is not continuous, it is only separately continuous--this is a general feature of non-normable locally convex spaces. Therefore, the preceding approach fails, which leads me to my question.

Question 1. Is there a way to define a "canonical" way to define a trace $\Tr$ on $\mathcal{S}(\mathbb{R}^k) \mathbin{\hat{\otimes}_\pi} \mathcal{S}'(\mathbb{R}^k)$ (i.e. a map such that $\Tr(f\otimes\bar{g}) = \overline{\langle{g,\bar{f}}\rangle}$)?

Question 2. If the answer to Question 1 is no, is there a non-canonical way of defining a trace $\Tr$ on $\mathcal{S}(\mathbb{R}^k) \mathbin{\hat{\otimes}_\pi} \mathcal{S}'(\mathbb{R}^k)$ in such a way that if $\gamma\in\mathcal{S}(\mathbb{R}^k) \mathbin{\hat{\otimes}_\pi} \mathcal{S}' (\mathbb{R}^k)$ and can be identified with an element of trace-class operators on $L^2(\mathbb{R}^k)$, then $\Tr$ coincides with the usual definition of trace?

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