Representation of a Schwartz map in terms of a kernel

Question

Suppose $f: \mathcal{S}(\mathbb{R}^{d})^{n+1} \to \mathbb{C}$ is a continuous function. To each $\varphi \in \mathcal{S}(\mathbb{R}^{d})$, we can define the map $f[\varphi]: \mathcal{S}(\mathbb{R}^{d})^{n} \to \mathbb{C}$ given by: \begin{eqnarray} f[\varphi](\psi_{1},...,\psi_{n}) := f(\varphi, \psi_{1},...,\psi_{n}). \tag{1}\label{1} \end{eqnarray} Note that $f[\varphi]$ is a continuous map. Now, let us assume that $f[\varphi]$ is also linear in each of its entries (or multilinear if you prefer). Then, $f[\varphi] \in \mathcal{L}(\mathcal{S}(\mathbb{R}^{d})^{n})$, where $\mathcal{L}(\mathcal{S}(\mathbb{R}^{d})^{n})$ denotes the space of all linear and continuous functions from $\mathcal{S}(\mathbb{R}^{d})^{n}$ to $\mathbb{C}$.

Let $\varphi \in \mathcal{S}(\mathbb{R}^{d})$ be fixed. I'd like to know if there exists some kernel $K_{\varphi} \in \mathcal{S}'(\mathbb{R}^{nd})$ such that \begin{eqnarray} K_{\varphi}(\psi_{1}\otimes\cdots\otimes \psi_{n})=f[\varphi](\psi_{1},...,\psi_{n}) \tag{2}\label{2} \end{eqnarray} where $(\psi_{1}\otimes \cdots \otimes \psi_{n})(x_{1},...,x_{n}) := \psi_{1}(x_{1})\cdots\psi_{n}(x_{n})$, for $\psi_{1},...,\psi_{n} \in \mathcal{S}(\mathbb{R}^{d})$. If $n=1$, I believe this has to do with the Schwartz Kernel Theorem, but I don't know how if the result follow for $n>1$. Does it follow by induction maybe?

EDIT: Just to clarify, I could have asked the question in a simpler way. The question is basically if, given a function $f \in \mathcal{L}(\mathcal{S}(\mathbb{R}^{d})^{n})$, there exists some kernel $K$ such that $K(\psi_{1}\otimes \cdots \otimes \psi_{n}) = f(\psi_{1},...,\psi_{n})$. I stated it differently because I'm thinking of $f[\varphi]$ to be a derivative $D^{n}f[\varphi]$, and this explains my initial notation.

Answer 1

A stronger assertion than what you ask for does hold. Namely, a continuous multilinear map of $S(\mathbb R^n) \times ... S(\mathbb R^n)$ ($d$ factors) to $\mathbb C$ is given by a tempered distribution on $\mathbb R^{nd}$.

First, the nuclearity of $S(\mathbb R^n)$ and $S(\mathbb R^{nd})$ (which is not a trivial thing) implies the existence of a genuine categorical tensor product of those Schwartz spaces, and further related specific computations show that that categorical tensor product is $S(\mathbb R^{nd})$... and then a continuous multilinear map to scalars does factor through a continuous linear map to scalars from that tensor product... which is a tempered distribution on $\mathbb R^{nd}$.

(If this response fails to address your issue, please advise...)

Answer 2

The proof of the Kernel Theorem for $\mathscr{S}$, $\mathscr{S}'$ is trivial modulo a nontrivial (but not so hard) theorem: the isomorphism with spaces of sequences, e.g., via Hermite functions which are the eigenvectors for the quantum harmonic oscillator. For a sketch of the proof of the KT see https://math.stackexchange.com/questions/3512357/understanding-the-proof-of-schwartz-kernel-theorem/3512932#3512932