I asked this question on Math StackExchange, but it did not receive an answer, despite my offering a bounty to attract attention. I am unsure whether it is appropriate for this venue, but I thought that I would try my luck. Below I have reproduced the question with some modifications.

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
$$\operatorname{Tr}(f\otimes \bar{g}) := \overline{\langle{g, \bar{f}}\rangle}_{\mathcal{S}'-\mathcal{S}},$$
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
$$\operatorname{Tr}: \mathcal{S}(\mathbb{R}^k) \mathbin{\hat{\otimes}_\pi} \mathcal{S}'(\mathbb{R}^k) \rightarrow \mathbb{C}$$
with the property that $\operatorname{Tr}(f\otimes \bar{g})$ 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 "canonical" way to define a trace $\operatorname{Tr}$ on $\mathcal{S}(\mathbb{R}^k) \mathbin{\hat{\otimes}_\pi} \mathcal{S}'(\mathbb{R}^k)$ (i.e. a linear map such that $\operatorname{Tr}(f\otimes\bar{g}) = \overline{\langle{g,\bar{f}}\rangle}$)?

It seems that such a map $\operatorname{Tr}$ cannot be continuous $\mathcal{S}(\mathbb{R}^k)\hat{\otimes}\mathcal{S}'(\mathbb{R}^k) \rightarrow \mathbb{C}$, otherwise, since the canonical bilinear map $$\mathcal{S}(\mathbb{R}^k)\times\mathcal{S}'(\mathbb{R}^k) \rightarrow \mathcal{S}(\mathbb{R}^k) \mathbin{\hat{\otimes}_\pi} \mathcal{S}' (\mathbb{R}^k), \qquad (f,g) \mapsto f\otimes g$$ is continuous, we would have the continuity of the evaluation map.

Question 2.If the answer to Question 1 is no, is there a non-canonical way of defining a trace $\operatorname{Tr}$ 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 $\operatorname{Tr}$ coincides with the usual definition of trace?