# Does the Fourier transform preserve the separation property?

The space of Schwartz functions on the plane is denoted by $\mathcal{S}$. The usual multiplication and the convolution multiplication on $\mathcal{S}$ are denoted by $m_1$ and $m_2$, respectively.

The Fourier transform $\mathcal{F}$ on $\mathcal{S}$ give a bijective correspondence between the $m_1$-subalgebras of $\mathcal{S}$ and $m_2$- subalgebras of $\mathcal{S}$.

We say that a subset $A$ of $\mathcal{S}$ separates compact subsets of $\mathbb{R}^2$ if for every disjoint compact set $K_1,K_2, \ldots,K_n$, there exist a function $f\in A$ such that $f(K_i) \cap f(K_j)$ is null, for $i\neq j$.

Is it true to say that the Fourier transform gives a bijective correspondence between separating $m_1$-subalgebras of $\mathcal{S}$ and its $m_2$ separating subalgebras?

The question is motivated by the following post:

Are these function spaces appropriate to be considered as the domain of certain differential operator?

• Do you know if the corresponding result is true one dimension lower, i.e. for Schwartz functions on the real line? Commented May 15, 2017 at 22:13

I can't give a precise or complete answer, but based on vague, spontaneous thoughts I suggest considering the subalgebra of the Schwartz space generated by all bump functions of the form $$f^c_x(y):=e^{-\frac{c}{\Vert x-y\Vert^2}},\quad x,y\in \mathbb{R}^2,\; c>0.$$ Clearly, this algebra is separating. In these notes it is shown that the Fourier transform $\hat f^1_x$ of $f^1_x$ fulfills $$\hat f^1_x(\xi)\sim 2 \mathrm{Re}\,\bigg(\sqrt{\frac{-i\pi}{\sqrt{2i}\xi^{3/2}}}e^{i\xi-1/2-\sqrt{2i\xi}}\bigg)\qquad \text{as }\xi \to +\infty.$$ Deducing from this the asymptotic behaviour of the Fourier transforms of the functions $f^c_x$ for $c\neq 1$, one should be able to conclude that the Fourier transforms $\hat f^c_x(\xi)$ oscillate very quickly as $\xi$ becomes large. This should imply that also their convolutions oscillate very quickly as $\xi$ becomes large, hence the algebra formed by the $\hat f^c_x(\xi)$ under convolution is not separating anymore because when one chooses two compact sets that are large enough and far enough away from the origin, then there won't be any two functions in the algebra that don't oscillate at least once inside each of the two compact sets.