# Representation of a Schwartz map in terms of a kernel

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.

• Just to be sure: $f$ is multilinear? – paul garrett Apr 8 at 21:26
• @paulgarrett $f[\varphi]$ is multilinear. As an example, take $f[\varphi]$ do be the $n$-th derivative of $f$ at $\varphi \in \mathcal{S}(\mathbb{R}^{d})$. – IamWill Apr 8 at 21:31
• So for each $\varphi$, $f[\varphi]$ is multilinear, but all we know about the full $f$ is that it's continuous? So you're really asking about an individual $f[\varphi]$, whose dependence on $\varphi$ is not really the point? I'm just slightly confused about the way you set things up... – paul garrett Apr 8 at 21:35
• Yes, you're correct. I think I should clarify this in the post, indeed. My relation (2) is just to stress that $f[\varphi]$ is continuous and linear for each $\varphi$ but my question is whether, for a fixed $\varphi$, there is such kernel $K_{\varphi}$. I'm really not interested in the dependence of $\varphi$. – IamWill Apr 8 at 21:44
• I edited it. I think it's better now. Thanks! – IamWill Apr 8 at 21:47

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 $$\mathbb R^{nd}$$.
• Thanks for the answer. Your first statement says that a continuous multilinear map of $\mathcal{S}(\mathbb{R}^{d}) \to \mathbb{C}$ is given by a tempered distribution on $\mathbb{R}^{nd}$. I've never found this result in books (only with $n=2$). Does it follow by induction on the case $n=2$? – IamWill Apr 9 at 0:32
• Yes, it would follow by induction, together with understanding that with a genuine tensor product, $S(\mathbb R^m)\otimes S(\mathbb R^n)\approx S(\mathbb R^{m+n})$. Then the associativity, etc., of (genuine, categorical) tensor products gives the result. – paul garrett Apr 9 at 1:31
• One can use induction but it is not necessary. The point is that the same proof for $n=2$ works for any $n$, but the notation is more painful so many authors just give the proof for $n=2$. – Abdelmalek Abdesselam Apr 9 at 14:14
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