# Is the Schwartz Function space separable?

Let the Schwartz function space on $\mathbb R^n$ be endowed with its usual metric structure. Is this space separable? Can anyone give a reference?

• I'd say the analytic Schwartz functions are dense. May 6, 2017 at 13:51
• Thanks! But what do you mean by analytic Schwartz functions? Do you mean real analytic ones or complex analytic ones? Could you give a link to the references? May 6, 2017 at 15:49
• $f \ast m^n e^{-\pi |m x|^2} \to f$ and it is (real) analytic May 6, 2017 at 15:57
• Thank you for your comment! So how is the set of analytic Schwartz functions related to the countability? May 6, 2017 at 16:32
• An analytic function is fully determined by the sequence of its derivatives at $0$. May 6, 2017 at 16:35

Schwartz space is separable as a metric space.

Indeed it is isomorphic to $$\mathfrak{s}$$, the space of sequences $$(x_n)_{n\ge 0}$$ of fast decay with topology defined by the seminorms $$||x||_k=\sup_n\ (1+n)^k |x_n|$$ In $$\mathfrak{s}$$ take almost finite sequences of rationals and that gives a countable dense subset.

• May I have a reference to your answer? Mar 4, 2019 at 11:10
• You mean, more precisely, a reference for the isomorphism of $S$ with $\mathfrak{s}$? In that case you can see aip.scitation.org/doi/abs/10.1063/1.1665472 but the proof is not very well written. Simon gave a much better presentation in section 6.4 on Hermite expansions of his book "Real Analysis : A Comprehensive Course in Analysis, Part 1" bookstore.ams.org/simon-1 Mar 4, 2019 at 13:59