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](https://en.wikipedia.org/wiki/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. 



Representation of a Schwartz map in terms of a kernel
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...)


