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?


2 Answers 2


Q1: No. You gave a proof above: $\overline{\langle f,\bar g\rangle} = \operatorname{Tr}(f\otimes g)$.

Q2: No, if $\operatorname{Tr}$ is supposed to be continuous. Namely, $\mathcal S(\mathbb R^k)\subset L^2(\mathbb R^k)$ continuosly, and $L^2(\mathbb R^k)\subset \mathcal S'(\mathbb R^k)$ countinuous and dense. So any continuous trace would immediately lead to a contradiction to Q1.


In the "classical theory of topological vector spaces" the questions like this are intricated (in my opinion, this is an artifical complexity, the Nature can't be so complicated). But in the theory of stereotype spaces they become simple: for each stereotype space $X$ (including ${\mathcal S}({\mathbb R}^k)$, since it is also stereotype as a Fréchet space) the pairing $$ (x,f)\in X\times X^\star\mapsto f(x)\in{\mathbb C} $$ is a continuous bilinear map in the stereotype sense, and, as a corollary, it can be extended to a continuous linear functional on the "projective stereotype tensor product" $\circledast$ (this is an analog of $\hat{\otimes}_\pi$ in the stereotype theory) $$ \operatorname{cont}: X\circledast X^\star\to {\mathbb C}. $$ This functional is called a "contraction", you can look at the details in my paper of 2003 (page 265).

If you want to define a trace for all operators $\varphi:X\to X$ that are images of the tensors $\alpha\in X\circledast X^\star$ under the Grothendieck transformation $X\circledast X^\star\to{\mathcal L}(X)$, then your space $X$ must have the stereotype approximation property. As far as I know, nobody was interested up to now, whether the space ${\mathcal S}({\mathbb R}^k)$ has the stereotype approximation, but at first glance this is true: one can try to use the same trick as I did in my paper of 2018 for the space ${\mathcal C}(G)$ of continuous functions on a locally compact group $G$. Also Albrecht Pietsch writes in his Nuclear locally convex spaces, 10.3.2, that the Hermite polynomials form a basis in the space ${\mathcal S}'({\mathbb R})$ (which coincides with the stereotype dual space ${\mathcal S}({\mathbb R})^\star$). I think something similar must be true for each ${\mathcal S}'({\mathbb R}^k)$ (with arbitrary $k\in{\mathbb N}$), and if so, then ${\mathcal S}'({\mathbb R}^k)={\mathcal S}({\mathbb R}^k)^\star$ and ${\mathcal S}({\mathbb R}^k)$ have the stereotype approximation property.


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