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Let $M$ be a $d$-dimensional (say, Hausdorff, paracompact, connected and oriented) smooth manifold, and $K\subset M$ compact with $\mathring{K}\neq\varnothing$. M. Valdivia has shown (based on previous results by himself and D. Vogt, see e.g. M. Valdivia, A representation of the space $\mathscr{D}(K)$, J. reine angew. Math. 320 (1980) 97-98) that the nuclear Fréchet space $\mathscr{D}(K)$ of smooth functions supported in $K$ is topologically isomorphic to the space $s$ of rapidly decreasing sequences: $$s=\{(a_n)_{n\in\mathbb{N}}\ |\ ((1+n)^k a_n)_{n\in\mathbb{N}}\text{ is bounded for all }k\in\mathbb{N}\}\ .$$ Let $\Phi:\mathscr{D}(K)\cong s$ denote a Valdivia-Vogt isomorphism (see Edit below). It is clear that the transpose ${}^t\Phi$ of $\Phi$ yields a topological isomorphism between the dual $s'$ of $s$ $$s'=\{(a_n)_{n\in\mathbb{N}}\ |\ ((1+n)^{-k}a_n)_{n\in\mathbb{N}}\text{ is bounded for some }k\in\mathbb{N}\}$$ and the dual $\mathscr{D}(K)'$ of $\mathscr{D}(K)$, which may be identified as a vector space with $\mathscr{D}'(\wedge^d T^*M\rightarrow M)/\mathscr{D}(K)^\perp$, where $$\mathscr{D}(K)^\perp=\{u\in\mathscr{D}'(\wedge^d T^*M\rightarrow M)\ |\ u(\varphi)=0\text{ for all }\varphi\in\mathscr{D}(K)\}$$ is the annihilator of $\mathscr{D}(K)$. It is clear that $\mathscr{D}(K)'$ contains $$\mathscr{E}'(K)=\{u\in\mathscr{E}'(\wedge^d T^*M\rightarrow M)\ |\ \text{supp }u\subset K\}$$ as a (closed) subspace (I apologize for the slightly unconventional notation). Since the sequences $e_j=(e_{j,n})_{n\in\mathbb{N}}$ given by $$e_{j,n}=\begin{cases} 0 & (n\neq j) \\ 1 & (n=j) \end{cases}$$ form a Schauder basis of both $s$ and $s'$, it is clear that $s$ is dense in $s'$.

Question: Is there a choice of $\Phi$ (see Edit below) such that it extends to a topological isomorphism between $\mathscr{E}'(K)$ and $s'$? Likewise, does the restriction of ${}^t\Phi$ to $s$ for such a $\Phi$ yield another topological isomorphism between $s$ and $\mathscr{D}(K)$?

My question is inspired by the known characterization of $\mathscr{D}([0,1])$ and $\mathscr{E}'([0,1])$ through the decay / growth of their Fourier coefficients in $[0,1]$.

Edit: As suggested by Jochen Wengenroth in the comments below, the recent works of Bargetz

  • C. Bargetz, Commutativity of the Valdivia-Vogt table of representations of function spaces. Math. Nachr. 287 (2014) 10-22
  • C. Bargetz, Explicit representations of spaces of smooth functions and distributions, J. Math. Anal. Appl. 424 (2015) 1491-1505
  • C. Bargetz, Completing the Valdivia-Vogt tables of sequence-space representations of spaces of smooth functions and distributions. Monatsh. Math. 177 (2015) 1-14

indicate that the answer to the question may be negative depending on the choice of Valdivia-Vogt isomorphism $\Phi$. More precisely, one needs first for ${}^t\Phi$ to map $\{e_j\ |\ j\in\mathbb{N}\}$ into $\mathscr{D}(K)$, which may not be the case.

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    $\begingroup$ Perhaps this article of Bargetz helps Completing the Valdivia-Vogt tables of sequence-space representations of spaces of smooth functions and distributions. Monatsh. Math. 177 (2015), no. 1, 1–14. $\endgroup$ Jan 18, 2019 at 11:19
  • $\begingroup$ @JochenWengenroth Thanks. I couldn't get access to this paper yet, but judging from other papers by the same author the commutative Valdivia-Vogt isomorphism table and its dual don't seem to merge into a single commutative table, which is essentially (a stronger version of) what I'm asking. $\endgroup$ Jan 18, 2019 at 11:39
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    $\begingroup$ I've managed to get access to the Monatsh. Math. paper of Bargetz you cited. Unfortunately, this paper, as its predecessors (C. Bargetz, Commutativity of the Valdivia-Vogt table of representations of function spaces, Math. Nachr. 287 (2014) 10-22; Explicit representations of spaces of smooth functions and distributions, J. Math. Anal. Appl. 424 (2015) 1491-1505), gets the dual Valdivia-Vogt structure table from the standard one by transposition, and not by extension as I asked for. Therefore, it's not obvious whether both embed into a common commutative structure table or not. $\endgroup$ Jan 18, 2019 at 14:28
  • $\begingroup$ Even if you look at the details, e.g. by looking at the explicit form of the common Schauder basis obtained in the J. Math. Anal. Appl. paper for each structure table (compare Prop. 3.2, pp. 1499 with Props. 4.1, 4.2 and the discussion in between them, pp. 1502-1503), restricting a transpose VV isomorphism to $s$ may not give me what I want, for the Schauder basis obtained for the dual table doesn't consist of smooth functions (they're integrable but singular at a single point). This means that one needs a specific choice of VV isomorphism to get a positve answer to my question. $\endgroup$ Jan 18, 2019 at 14:54
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    $\begingroup$ I agree that the results you quoted do not provide a positive answer to your question. I believe that the question of whether both diagrams can be combined into one commutative diagram is open. $\endgroup$
    – Christian
    Jan 18, 2019 at 17:16

2 Answers 2

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This is just an addendum to the above answers so really a comment but too long--I hope that it will be of interest. At first, there is an abstract construction which associates to each self adjoint, unbounded operator on Hilbert space in a functorial way a densely embedded Fréchet space, whose dual space os identifiable as a superspace. The point is that if we take some of the classical such operators (notably, Laplace-Beltrami and Schrödinger operators), one obtains a unified approach to many (most?) of the standard spaces of distributions. Before giving example, let me note that the relevance to your question lies in the fact that if the operator has a discrete spectrum with eigenvalues that are asymptotically like a positive power of $n$, then we have precisely your situation--the ONB of eigenvectors is also a basis for the two derived spaces a d this establishes there isomorphism with $s$ and $s'$.

The canonical example is that of the Laplace-Beltrami operator on a closed compact Riemannian manifold where one gets the spaces of smooth functions and distributions thereon. The most important cases are the hypersphere and the higher dimensional tori. One can also consider the Laplacian on sufficiently regular open subsets of eucliidean space, with suitable boundary conditions (Dirichlet, Neumann, mixed). In the case of the distributions on the closed unit interval, one uses the Legendre differential operator with the Legendre function as basis. Many concrete compact subsets of euclidean space have been studied in detail (more recently, the case of fractals) but as mentioned above there doesn't seem to be a theory which covers your question in full generality.

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  • $\begingroup$ I suppose you are talking about the Gel'fand triple associated to an unbounded self-adjoint operator... However, as far as I know, the examples of Valdivia-Vogt isomorphisms for (say) $\mathscr{D}(K)$ are usually unrelated to spectral theory, even because the boundary of $K$ can be pretty much anything (besides being compact, of course). In fact, I've never seen an example of a Gel'fand triple where the original space is (say) $L^2(K,d\mu_g)$, with $d\mu_g$ a volume element associated to some Riemannian metric on $M$, and the "small" (i.e. "smooth") subspace is $\mathscr{D}(K)$ for any $K$. $\endgroup$ Feb 3, 2021 at 18:55
  • $\begingroup$ We seem to be talking at cross purposes. There are millions of such examples--try the case of hyperspheres for a start. The one dimensional case produces the periodic distribution with the trigonometric functions as basis. The case of the unit interval comes from the Legendre equation and has the corresponding eigenvectors as basis. $\endgroup$ Feb 4, 2021 at 5:45
  • $\begingroup$ Let me be more specific. I do know about specific cases of the region $K$ for which Gel'fand triples can be built. My concern about this method has two preliminary aspects: (a) does it recover precisely the space of smooth funcions on $M$ and compactly supported in $K$ if $\partial K\neq\varnothing$? (it certainly does for empty boundaries) (b) does it work with arbitrarily rough boundaries? If one can answer "yes" to both, one can then address the main point of my question using this framework: does the basis of eigenfunctions provide a Schauder basis in the form required in the OP? $\endgroup$ Feb 4, 2021 at 6:06
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Not a full answer but too long for am comment. For certain compact sets the answer might be positive when the isomorphism given by the Wilson basis constructed in the following paper is used:

C. Bargetz, A. Debrouwere, E. A. Nigsch: Sequence space representations for spaces of smooth functions and distributions via Wilson bases. arXiv: 2107.00245

In particular the isomorphisms there have the property that all of them arise from a single isomorphism between $\mathscr{D}'(\mathbb{R}^{n})$ and $s'\widehat{\otimes}\mathbb{C}^{\mathbb{N}}$.

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