$\mathcal S'(\mathbb R^d)$ is separable [closed]

Question

I Think the statement is true, but I struggle to find a reference for the fact that the space of tempered distributions equipped with the weak-* topology is separable. Thank you for your help!

Answer 1

If you mean by separable the existence of a countable dense subset the answer is yes. One can see this as follows. Let $\mathcal{s}$ be the space of sequences $x=(x_n)_{n\in\mathbb{N}}$ of real numbers for which all the seminorms $$ ||x||_{k}=\sup_{n\ge 0} (n+1)^k|x_n|\ , $$ $k\in\mathbb{N}$, are finite (I am using Bourbaki's convention of counting from zero). Equip this space with the locally convex topology generated by the countable collection of seminorms $||\cdot||_k$, $k\ge 0$. Let $\mathcal{s}'$ be the topological dual equipped with the weak-$\ast$ topology. Then $S'(\mathbb{R}^d)$ is isomorphic to $\mathcal{s}'$ and therefore your question can be reduced to the much more concrete case of the space $\mathcal{s}'$. The latter can (even more concretely!) be realized as the space of sequences $y=(y_n)_{n\in\mathbb{N}}$ of real numbers such that there exists a $k\ge 0$ and a constant $C$ for which $$ |y_n|\le C(n+1)^k $$ for all $n$. The duality pairing is given by $$ y(x)=\sum_{n=0}^{\infty} y_n x_n\ . $$ Now let $D=\oplus_{\mathbb{N}} \mathbb{Q}\subset \mathcal{s}'$, namely the set of almost finite sequences of rational numbers. $D$ is countable and dense.

Indeed let $y\in\mathcal{s}'$, let $x^{(1)},\ldots,x^{(p)}$ be in $\mathcal{s}$ and let $\epsilon>0$. Clearly, there exists a $z\in D$ such that for all $q$, $1\le q\le p$, $$ |(z-y)(x^{(q)})|<\epsilon\ . $$ Since the series defining the $y(x^{(q)})$ converge, truncate them far enough and approximate the corresponding finite sequence of terms in $y$ by rationals.

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Addendum as per Eric's comment: the isomorphism between $S'(\mathbb{R}^d)$ and $\mathcal{s}'$ is very well explained (using Hermite functions) in the article "Distributions and Their Hermite Expansions" by Barry Simon. It is funny how little known this thing is. I think also it is an open problem to find an explicit isomorphism (or Schauder basis) between $\mathcal{D}(\Omega)$ and $\oplus_{\mathbb{N}} \mathcal{s}$, see this other MO question.

Answer 2

Very generally, any dual of a space of continuous functions on a separable space is weak*-separable. More precisely, let $A$ be any topological space, let $B\subseteq A$ be a dense subset, and let $S$ be any vector space of continuous real-valued functions on $A$. Then I claim that $\mathbb{Q}$-linear combinations of evaluations $\delta_b$ at points $b\in B$ are dense in the full linear dual $S'$ with the weak* topology (and thus also in the continuous dual for any topology on $S$ that makes each $\delta_b$ continuous).

To show this, let $U\subset S'$ be a nonempty open set; then $U$ contains a subset of the form $$\{\alpha\in S': \alpha(f_1)\in V_1, \alpha(f_2)\in V_2,\dots,\alpha(f_n)\in V_n\}$$ where $f_1,\dots f_n\in S$ are linearly independent functions and $V_1,\dots,V_n\subset\mathbb{R}$ are nonempty and open. Since $B$ is dense in $A$, we can find points $b_1,\dots,b_n\in B$ such that the restrictions of the $f_i$ to $\{b_1,\dots,b_n\}$ are still linearly independent (prove this by induction on $n$). We thus have a perfect pairing between the span of the $f_i$ and the span of the functionals $\delta_{b_i}$. Since the $V_i$ are open, it follows that we can find a $\mathbb{Q}$-linear combination of the $\delta_{b_i}$ which is in our $U$.