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The more I study Schwartz distributions and the corresponding spaces, the more the latter look "finite dimensional" to me. Of course they are not finite dimensional in the technical sense but they are essentially finite dimensional in the sense that "theorems which are true for $\mathbb{R}^n$ should be true for them too".

I would like to get a better hold on what "essentially finite dimensional" should mean. So I propose this question/challenge (maybe this should be made community wiki?).

Question: Can you find examples of theorems which only use the topological vector space structure of $V$ and are true when $V$ is finite dimensional, typically false when $V$ is an infinite dimensional Banach space, but true again when $V$ is $\mathcal{S}'(\mathbb{R}^n)$ or $\mathcal{D}'(\mathbb{R}^n)$ (with the strong topology).

Here are some example to get started.

  1. A subset $A$ of $V$ is compact iff it is closed and bounded.
  2. $V$ is reflexive.
  3. A sequence of Borel probability measures on $V$ converges weakly iff the characteristic functions $V'\rightarrow\mathbb{C}$ converge pointwise to a function that is continuous at the origin.

I would prefer examples of properties/theorems which are not entirely a consequence of being nuclear.


Motivation: I don't yet have a definition of "essentially finite dimensional" (EFD) which is a fluctuating concept to me. Of course EFD should be of the form "nuclear+some mystery ingredients". I am hoping the answers to the question as I asked it would tell me more about the optimal mix of mystery ingredients rather than the nuclear part. One might think that what I am after is a phrase-definition like "nuclear, barrelled and quasi-complete" but it is not quite that as I try to explain in what follows.

The definition of finite dimensional vector space is that $V$ is isomorphic to $\mathbb{R}^n$ for some $n$. This involves a concrete model as a space of functions on a discrete set, here the finite set $\{1,2,\ldots,n\}$. I suppose EFD could be defined similarly, as follows. Let $\Omega$ be a set of "pre-seminorms" on $\mathbb{R}^{\mathbb{N}}$. By "pre-seminorm" I mean the same definition as seminorm but allowing the value $\infty$. Then define $V=\{x\in \mathbb{R}^{\mathbb{N}} | \forall \omega\in\Omega, \omega(x)<\infty\}$ with the topology defined by the restrictions of the elements of $\Omega$ which then become true seminorms. I suppose one could restrict to $l^1$ pre-seminorms of the form $$ \omega(x)=\sum_{n\in \mathbb{N}}\omega_n\ |x_n| $$ with weights $\omega_n\ge 0$. What I would ultimately like is an explicit property of $\Omega$ which makes the corresponding space as close as possible to a finite dimensional space. The part of that property concerning nuclearity was worked out at long time ago by Grothendieck for echelon spaces and Pietsch in general. The property should be "The Pietsch criterion+???".

Anyway, all I asked is examples of theorems, the more surprising the better, and the further away from just being nuclear the better too.


Motivation for the motivation: I have been thinking about how one could improve existing presentations of the theory of distributions, see e.g. Nice applications for Schwartz distributions One approach I am exploring is to use sequence space representations and I would eventually like to narrow down the possibilities to a class of $\Omega$'s which is easy to define and could serve as "optimal" class of concrete models for EFDs.

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    $\begingroup$ Nuclear space? $\endgroup$ – Mateusz Kwaśnicki Apr 6 '18 at 19:04
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    $\begingroup$ Surely there is a more polite way to indicate that a reasonable answer is not the sort you want than to say that it "is completely off the point". (It doesn't seem fair to say that @MateuszKwaśnicki was reiterating, since your post doesn't directly mention nuclear spaces ….) $\endgroup$ – LSpice Apr 6 '18 at 20:12
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    $\begingroup$ @AbdelmalekAbdesselam: By no means I intended to offend you. I just wanted to point out that the properties that you listed (as well as, for example, Schwartz kernel theorem and Bochner's theorem, listed in the linked Wikipedia entry) are specific to nuclear spaces. If you are looking for other properties, you may want to explain in what sense your concept of essentially finite-dimensional spaces is supposed to be different from the notion of nuclear spaces. $\endgroup$ – Mateusz Kwaśnicki Apr 7 '18 at 8:29
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    $\begingroup$ @AbdelmalekAbdesselam I see no need for an apology. You're seeing a personal affront where none was intended. In fact, Mateusz was moments faster before I would have posted a similar (but longer) comment. Nuclearity is an obvious first thought here and I (who don't know you or the amount of your knowledge) cannot see any reason not to cover the obvious thoughts first, before we move on to different approaches. I agree with Mateusz on the point that it may be helpful to explicate exactly what differences to nuclearity you want to have... $\endgroup$ – Johannes Hahn Apr 7 '18 at 17:52
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    $\begingroup$ Storm in a teacup? $\endgroup$ – Liviu Nicolaescu Apr 8 '18 at 22:18
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Like others have pointed out the key, concept is that of nuclear spaces and a good presentation can be found in volume 4 of Gelfand's Generalized Functions.

Kolmogorov has introduced a concept of dimension of a functional space and the nuclear spaces have finite Kolmogorov dimension; see Chap I. Sec. 3.8 of the above reference. For example, in this section it is proved the space of real analytic functions on $\mathbb{R}^n$ has Kolmogorov dimension $n$. As an aside, this concept of dimension has lead to proofs of many generalizations of Hilbert 13th problem.

The above reference also contains a beautiful presentation of the Schwartz kernel theorem. I recommend this source enthusiastically.

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  • $\begingroup$ Thank you Liviu. I looked at bits and pieces of the Generalized Functions series, but that was a while ago. I also share your high esteem for this work, see for example what I said at the end of arxiv.org/abs/1511.03180 My (perhaps faulty) recollection was that they only considered the countably Hilbert particular case of Grothendieck's nuclear property. In any case your answer prompts me to go have a second look at the books. $\endgroup$ – Abdelmalek Abdesselam Apr 8 '18 at 15:34
  • $\begingroup$ Yes, it is true that they cover only countable Hilbert spaces but this covers the spaces needed in distribution theory. $\endgroup$ – Liviu Nicolaescu Apr 8 '18 at 16:56
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Every short exact sequence $$0\to \mathscr S'\to Y\to \mathscr S'\to 0$$ (where $Y$ is any locally convex space) splits, i.e., the quotient map $Y\to \mathscr S'$ has a continuous linear right inverse. The same for $$0\to \mathscr D'\to Y\to \mathscr D'\to 0$$ where $Y$ is any so-called PLS-space. These are very particular cases of the splitting theory for Frechet spaces (plus duality) and PLS-spaces developped by Palamodov and Vogt (and others, in particular, Wagner, Domanski,...). These results use more than the forbidden property.

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  • $\begingroup$ Thank you Jochen. This is the kind of answers I was hoping for. There is no forbidden property. It's just that I prefer theorems which do not only rely on the nuclear property. $\endgroup$ – Abdelmalek Abdesselam Apr 8 '18 at 14:37
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The Schwartz kernel theorem: For a manifold $M$ we have \begin{multline*} \mathcal D'(M\times M) = (\mathcal D(M)\hat{\hat\otimes} \mathcal D(M))' = (\mathcal D(M)\hat\otimes \mathcal D(M))' = L(\mathcal D(M), \mathcal D'(M))\,. \end{multline*} where one needs first the completed inductive or $\epsilon$-tensorproduct, and then the projective one. Fortunately they agree by who shall not be named, just like in finite dimensions.

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  • $\begingroup$ The injective tensor product of the space of test functions on the real line with itself isn't the space of test functions on the plane! $\endgroup$ – Jochen Wengenroth Apr 7 '18 at 15:35
  • $\begingroup$ @Peter: Thank you! I suppose the first equality uses more than the TVS structure and so does not fit the format of my question, but the last one does. $\endgroup$ – Abdelmalek Abdesselam Apr 7 '18 at 15:47

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