Trace on $\mathcal{S}(\mathbb{R}^k) \mathbin{\hat{\otimes}_\pi} \mathcal{S}'(\mathbb{R}^k)$

$$\newcommand{\Tr}{\operatorname{Tr}}$$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?