# Schwartz space on $\bigcup_{n=1}^CR^n$

I have an application where I need to work with the following idea. Let the space $\bigcup_{n=1}^C \mathbb{R}^n$ be associated with the metric $d$ such that for $x=(x_1,\cdots,x_n)$ and $y=(y_1,\cdots,y_m)$, we have $d(x,y) = \begin{cases} \sum_{i=1}^n\vert{x_i-y_i}\vert &\text{if } n=m, \\ \vert{m-n}\vert & \text{Otherwise. } \end{cases}$

A function $\phi$ on $\bigcup_{n=1}^C \mathbb{R}^n$ is a collection of functions $\{\phi_n, 1\leq n\leq C\}$, where $\phi_n$ is defined on $\mathbb{R}^n$.Similarly, let $\nu$ be a finite measure defined on $\bigcup_{n=1}^C \mathbb{R}^n$.

I need to work with the Schwartz space on $\bigcup_{n=1}^C \mathbb R^n$ and its dual by defining an inner product as $\langle\nu,\phi\rangle=\sum_{n=1}^C\int_{\mathbb R^n}\phi(x_1,\dotsc,x_n)\ d\nu(x_1,\dotsc,x_n)$. Do all the properties of Schwartz space and its dual on just $\mathbb R^n$ holds even for the case of the space $\bigcup_{n=1}^C\mathbb R^n$? If it is true, can we extend this to the case when $C=\infty$? Please give some suggestions and references for this.

I work on the stochastic interacting particle systems containing $N$ particles in which each particle state lies in the space $\bigcup_{n=1}^C \mathbb R^n$. The counting measure $\nu$ on $\bigcup_{n=1}^C \mathbb R^n$ can be considered as a collection of $C$ counting measures $\nu=(\nu_n,1\leq n\leq C)$ where $\nu_n$ for $1\leq n\leq C$ is a counting measure on $\mathbb{R}^n$. By considering $\nu$ lies in the dual of a Schwartz space, I want to establish the functional central limit theorem by taking $N\to\infty$ using the Mitoma's theorem. In general, the results are known for the case when $\nu$ lies on just $\mathbb{R}^n$. In my case, I can view $\nu$ as a single measure on $\bigcup_{n=1}^C \mathbb R^n$ or a collection of measures $(\nu_n,1\leq n\leq C)$ where $\nu_n$ lies on $\mathbb{R}^n$. If I view $\nu$ as collection of measures, a sequence of measures $\{\eta^m\}$ converges to $\eta$ if $\{\eta^m_i\}$ converges to $\eta_i$ for all $1\leq i\leq C$. I am looking at the notion of the convergence of the measures on $\bigcup_{n=1}^C \mathbb R^n$ by considering they lie in the dual of a Schwartz space. Please let me know any useful references or suggestions.

• Is that union disjoint? (otherwise, for finite $C$, I don't see any difference with $\mathbb{R}^C$) – Qfwfq Aug 8 '18 at 16:49
• You should explain what you mean by $\cup_{n=1}^C{\mathbb R}^n$. – Sergei Akbarov Aug 8 '18 at 16:50
• Naive question trying to understand your post: if I write $\mathcal{S}_n$ for the Schwartz space on $\mathbb{R}^n$, is your "Schwartz space on $\cup_{n = 1}^C \mathbb{R}^n$" just $\oplus_{n = 1}^C \mathcal{S}_n$? Or is it something else? – Willie Wong Aug 8 '18 at 19:05
• I would also like to see some clarification on the sentence "I need to work with the Schwartz space on $\bigcup_n \mathbb{R}^n$". Why exactly do you need to do that? Taking a setwise union of vector spaces is something so weird that I would like to know what drove you to it. I mean, even the types of the elements don't match. Why would anyone ever want to unionise these spaces? Are you sure you're not trying to do something completely different here? – Johannes Hahn Aug 8 '18 at 22:19
• I think the OP really means disjoint union of the type $\cup_n X^n$. This is not uncommon especially with interacting particle system picture in the background. The linear structure of $X$ plays no role. – Abdelmalek Abdesselam Aug 9 '18 at 20:09

Let $\mathscr{S}(\mathbb{R}^d)$ denote the usual Schwartz space of smooth rapidly decaying functions on $\mathbb{R}^d$. As suggested in Willie Wong's comment it seems the wanted space is just $S=\oplus_{n=1}^{C}\mathscr{S}(\mathbb{R}^n)=\prod_{n=1}^{C}\mathscr{S}(\mathbb{R}^n)$ with the product topology. It turns out, as topological vector spaces, $S\simeq\mathscr{S}(\mathbb{R})$. This follows from the Hermite representation, see, e.g., this article by Simon (there is a nicer presentation in his new comprehensive course on analysis).
Let $\langle x\rangle=\sqrt{1+\sum_{i=1}^{d}x_i^2}$ for $x\in \mathbb{R}^d$ and in particular also for multiindices $\alpha\in\mathbb{N}_0^{d}\subset\mathbb{R}^d$. For $k\in\mathbb{N}_0$ and for a (multi)sequence $z=(z_{\alpha})_{\alpha\in\mathbb{N}_0^d}$ define $$||z||_{k}=\sup_{\alpha\in\mathbb{N}_0^d} \langle \alpha\rangle^k |z_{\alpha}| \in [0,\infty]\ .$$ Let $\mathcal{s}(\mathbb{N}_0^{d})$ be the space of $z$'s such that $\forall k, ||z||_k<\infty$ with the topology defined by these seminorms. Then the Hermite representation says that $\mathscr{S}(\mathbb{R}^d)\simeq\mathcal{s}(\mathbb{N}_0^{d})$ (the Hermite functions or eigenstates of the $d$-dimensional harmonic oscillator form an unconditional Schauder basis of $\mathscr{S}(\mathbb{R}^d))$. By a (polynomially bounded) relabeling, $\mathcal{s}(\mathbb{N}_0^{d})\simeq\mathcal{s}(\mathbb{N}_0)$. So $$S\simeq \prod_{n=1}^{C} \mathcal{s}(\mathbb{N}_0)\simeq \mathcal{s}(\mathbb{N}_0^{d})$$ where the last isomorphism follows by interlacing.
The dual $S'$ is well-defined. My (wild) guess is that the OP is interested in a limit theorem of probability (like the CLT) on $S'$. This is about the weak convergence of Borel probability measures on $S'$ with the strong topology. By a theorem of Fernique this amounts to showing pointwise convergence of characteristic functions to a function on $S$ which is continuous at the origin. Tightness (I suppose that's why the OP refers to Mitoma's Theorem) is also equivalent to equicontinuity of characteristic functions at the origin. A recent account of such topics is in this article.
This also works for $C=\infty$ but one has at least two choices. One option is to take the Borchers algebra $$S=\oplus_{n=0}^{\infty}\mathscr{S}(\mathbb{R}^n)$$ with the topology defined by all seminorms which are continuous on individual summands. Then $S'$ is isomorphic to $\mathscr{D}'(\mathbb{R})$ the space of (not necessarily temperate) distributions.
Another is to take for $S$ a space of sequences of sequences $z^{(m)}$ with fast decay in the $m$ direction too. But this is the same as $\mathcal{s}(\mathbb{N}_0^2)$.