4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ A special case which I think should be relatively easy is that of quadratic translation-invariant norms, of the form $||u||^2=\int|\hat u(\xi)|^2\rho(\xi)\ d\xi$ $\endgroup$ – Jean Duchon Jan 27 '18 at 17:58
  • $\begingroup$ Did you lookup the related notions of "space of distributions" and "normal space of distributions"? $\endgroup$ – Abdelmalek Abdesselam Jan 27 '18 at 21:59
  • $\begingroup$ 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 $X$ ? $\endgroup$ – Johannes Hahn Jan 27 '18 at 22:40
  • $\begingroup$ 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. $\endgroup$ – paul garrett Jan 27 '18 at 22:53
  • 2
    $\begingroup$ 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 $f\in\mathcal S\}$. $\endgroup$ – Jochen Wengenroth Jan 29 '18 at 11:38
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – Goulifet Feb 2 '18 at 15:53
  • $\begingroup$ 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? $\endgroup$ – Goulifet Feb 2 '18 at 15:55
  • $\begingroup$ @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... $\endgroup$ – Sergei Akbarov Feb 3 '18 at 6:44
0
$\begingroup$

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 $D^{-1}L^2/P_0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.