# Completion of $\mathcal{S}(\mathbb{R})$ for a given norm

Assume that $\lVert \cdot \rVert$ is a norm on the space of rapidly decaying functions $\mathcal{S}(\mathbb{R})$. Under which conditions on the norm can we say that the completion $\mathcal{X}$ for this norm is a Banach space such that \begin{equation} \mathcal{S}(\mathbb{R}) \subseteq \mathcal{X} \subseteq \mathcal{S}'(\mathbb{R}), \end{equation} where $A \subseteq B$ means that the topological vector space $A$ is continuously embedded in the topological vector space $B$. This is typically valid for the norm $\lVert \cdot \rVert_{L^2}$ but not for the norm $\lVert \mathrm{D} \cdot \rVert_{L^2}$, where $\mathrm{D}$ is the derivative operator.

NB. Here, the space of tempered generalized functions $\mathcal{S}'(\mathbb{R})$ is endowed with the weak* topology.

• A special case which I think should be relatively easy is that of quadratic translation-invariant norms, of the form $||u||^2=\int|\hat u(\xi)|^2\rho(\xi)\ d\xi$ – Jean Duchon Jan 27 '18 at 17:58
• Did you lookup the related notions of "space of distributions" and "normal space of distributions"? – Abdelmalek Abdesselam Jan 27 '18 at 21:59
• And what kind of inclusion do you want to look at? I assume you want $\mathcal{S}\subseteq X$ to be the natural inclusion into the completion. What map do you want for $X\subseteq \mathcal{S}'$ ? Usually one would look at Hilbert spaces and consider this as the dual of the other embedding while identifying $X$ with $X'$ via the scalar product. What do you want to do for a general Banach space $X$ ? – Johannes Hahn Jan 27 '18 at 22:40
• Continuing @JohannesHahn's comment, if $\mathcal S\subset X$ is continuous, and if there is a natural, continuous $X\subset X'$, then the dual/adjoint of the inclusion $\mathcal S\subset X$ is $X'\subset \mathcal S'$, and splicing all these together gives the picture you want. – paul garrett Jan 27 '18 at 22:53
• Grothendieck's factorization theorem gives a necessary condition for the continuity of $j:X\to \mathcal S'$, namely that it factorizes over a "step" $X_n=\{u\in \mathcal S': |u(f)|\le c \sup\{(1+|x|^2)^n |f^{(\alpha)}(x)|: x\in\mathbb R^N,|\alpha|\le N\}$ for some $c\ge 0$ and all $f\in\mathcal S\}$. – Jochen Wengenroth Jan 29 '18 at 11:38

I am not sure that this is what you want, but it's too long for a comment, so I post it as an answer.

I don't see serious problems. If the conditions in terms of the theory of topological vector spaces will satisfy you, then you can consider the following. Your norm $\|\cdot\|$ must

1. be continuous on ${\mathcal S}({\mathbb R})$ (in other words, the unit ball $B$ of this norm must be a neighbourhood of zero in ${\mathcal S}({\mathbb R})$), and

2. generate a topology on ${\mathcal S}({\mathbb R})$ which is stronger than the weak topology generated on ${\mathcal S}({\mathbb R})$ by the duality $$\langle f,g\rangle=\int_{\mathbb R}f(t)\cdot g(t)\, d t,\quad f,g\in{\mathcal S}({\mathbb R})$$ (in other words, $\forall g\in{\mathcal S}({\mathbb R})\ \sup_{f\in{\mathcal S}({\mathbb R}):\ \|f\|\le 1}|\langle f,g\rangle|<\infty$).

If you denote by ${\mathcal S}({\mathbb R})_{\|\cdot\|}$ the space ${\mathcal S}({\mathbb R})$ endowed with the topology generated by such a norm, then you obtain a chain of continuous mappings $${\mathcal S}({\mathbb R})\subseteq{\mathcal S}({\mathbb R})_{\|\cdot\|}\subseteq {\mathcal S}'({\mathbb R}),$$ and the completion turns it into the chain $${\mathcal S}({\mathbb R})\subseteq X\to {\mathcal S}'({\mathbb R}).$$ (since ${\mathcal S}({\mathbb R})_{\|\cdot\|}$ is normed, its completion consists of Cauchy sequences; on the other hand, by the Banach-Steinhaus theorem, ${\mathcal S}'({\mathbb R})$ is sequentially complete with respect to the ${\mathcal S}({\mathbb R})$-weak topology, so the images of these sequences have limits in ${\mathcal S}'({\mathbb R})$).

The only detail which is not clear here is if the last mapping (the arrow $\to$) is injective. In general case completion does not preserve injectivity, so you should verify this in case that this is important for you.

You can also consider the strong topology on ${\mathcal S}'({\mathbb R})$ (i.e. the topology of uniform convergence on bounded or, what is the same here, on totally bounded sets in ${\mathcal S}({\mathbb R})$), where the same reasoning works also.

• Your necessary requirement on $\lVert \cdot \rVert$ are interesting. Especially, the point 2. is what does not work with the norm $\lvert f' \rVert_{L^2}$, for which $\sup_{f \in \mathcal{S}, \lVert f' \rVert_{L^2}} \lvert \langle f , g \rangle \rvert = \infty$ for most $g\in \mathcal{S}$ (as soon as the mean of $g$ is nonzero, I guess). – Goulifet Feb 2 '18 at 15:53
• I should also say that the injectivity is important to me. My goal is to identify when the completion $X$ is a "valid" function space, hence I really want elements in $X$ to be distinct elements in $\mathcal{S}'$. Do you see any additional condition preserving the injectivity? – Goulifet Feb 2 '18 at 15:55
• @Goulifet , it doesn't come to mind immediately, why (or when) this mapping has to be injective... Maybe I'll think a bit more, and later tell something clever on that score... – Sergei Akbarov Feb 3 '18 at 6:44

In the general case of a normed linear space $X$ and a larger quasi-complete Hausdorff topological vector space $E$ (as $\mathcal S'$ is) with $X\subseteq E$ : it can be completed within $E$ iff its unit ball is closed for the topology induced on $X$ by $E$.

While this is a basic fact, I'm not sure it is present in textbooks. I found it in Laurent Schwartz's article Sous-espaces hilbertiens d'espaces vectoriels topologiques et noyaux associés (Noyaux reproduisants), 1964. The proof is for quadratic norms, but is clearly valid for any norms. An interesting counter-example he gives (p.130) is the squared norm $\int \psi^2+\psi(0)^2$ for which the completion is $\{(f,\alpha):f\in L^2,\alpha\in \mathbb R\}$, not a subspace of $E=\mathcal S'$.

For (squared) norms of the special form $\int_{\mathbb R^d}|\hat u(\xi)|^2|\xi|^{2s}\ d\xi$, the condition is $-\frac{d}2<s<\frac{d}2$. If $s\ge\frac{d}2$ the completion is a quotient, a Hilbert subspace of $\mathcal S'/P$ where $p\in P$ iff there is a sequence $\psi_n\to p$ (in $\mathcal S'$) s.t. $||\psi_n||\to 0$. In your case ($d=1$, $s=1$), $P$ is just the space of constant functions. (Idem in the critical case $s=\frac12$, but that's not as easy then). Your completion is the Hilbert space I would denote $D^{-1}L^2/P_0$.