# Is the 3d Schwartz space isomorphic to a subspace of the 1d Schwartz space?

Are the Schwartz-spaces $\mathscr{S}(\mathbb{R})$ and $\mathscr{S}(\mathbb{R}^3)$ isomorphic (as topological vector spaces)? Is $\mathscr{S}(\mathbb{R}^3)$ at least isomorphic to a subspace of $\mathscr{S}(\mathbb{R})$? (These are simplified versions of item 3. and 4. from this question. Sorry for reposting them here...) I think that $\mathscr{S}(\mathbb{R})$ is isomorphic to a subspace of $\mathscr{S}(\mathbb{R}^3)$, for example to the subspace of functions of the form $f(x)\exp(-y^2-z^2)$. Is this correct?

One hint that $\mathscr{S}(\mathbb{R}^3)$ might indeed be isomorphic to a subspace of $\mathscr{S}(\mathbb{R})$ can be found in the Encyclopedia of Mathematics article Nuclear space:

Every nuclear space of type Fréchet is isomorphic to a subspace of the space $\mathscr{E}(\mathbb R)$ of infinitely-differentiable functions on the real line, that is, $\mathscr{E}(\mathbb R)$ is a universal space for the nuclear spaces of type Fréchet (see [10]).
[10] T. Komura, Y. Komura, "Ueber die Einbettung der nuklearen Räume in $(s)^A$" Math. Ann. , 162 (1965–1966) pp. 284–288

• "isomorphism" = "isomorphism of topological vector spaces"? – Yurii Savchuk Feb 16 '15 at 10:55

## 1 Answer

Let $\{e_k,\ k\in\mathbb N_0\}$ and $\{e_{ijk},\ (i,j,k)\in\mathbb N_0^3\}$ be the ONBs consisting of Hermitian functions in $L^2(\mathbb R)$ and $L^2(\mathbb R^3)$ respectively. We want to construct an isomorphism which maps one ONB bijectively onto another ONB.

Let $\varphi:\mathbb N_0^3\to\mathbb N_0$ be the enumerating function (which is a bijection). Then we have a Hilbert space isomorphism $$\Phi:L^2(\mathbb R^3)\to L^2(\mathbb R)$$ defined by $e_{ijk}\mapsto e_{\varphi(i,j,k)}.$ Some easy combinatorics shows that $(i+j+k)\leq\varphi(i,j,k)\leq c(i+j+k)^3, c>0.$

Using the $N$-presentation for $\mathscr{S}$ (see e.g. the book by Reed, Simon vol.1) we have $$\sum\lambda_{ijk}e_{ijk}\in\mathscr{S}(\mathbb{R}^3)\Longleftrightarrow \sup_n|\lambda_{ijk}|^2\cdot|i+j+k|^n<\infty\\ \Longleftrightarrow \sup_n|\lambda_{ijk}|^2\cdot|i+j+k|^{3n}<\infty,\forall n\in\mathbb N_0\\ \Longleftrightarrow\sup_n|\lambda_{ijk}|^2\cdot|\varphi(i,j,k)|^n<\infty\Longrightarrow \sum\lambda_{ijk}e_{\varphi(i,j,k)}\in\mathscr{S}(\mathbb{R})$$

which shows that $\Phi$ is a linear bijection of $\mathscr{S}(\mathbb{R}^3)$ onto $\mathscr{S}(\mathbb{R}).$

You can show that $\Phi,\Phi^{-1}$ are continuous by considering the family of seminorms $$||x||_n=||(N+1)^nx||_{L^2},$$ where $N$ is the number operator $Ne_k=ke_k$ resp. $Ne_{ijk}=(i+j+k)e_{ijk}.$ Hence $\Phi$ is an isomorphism.