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Part of me is afraid that this isn't a well-formed question, but try as I might, I can't seem to figure out anything reasonable on this topic. I'm hoping someone here can help.

In functional analysis, one defines the Schwartz space (on $\mathbb{R}^n$, say) as the collection of all smooth functions whose derivatives decay "rapidly" (in the very precise sense given, e.g., on Wikipedia). Over at the nLab (http://ncatlab.org/nlab/show/Schwartz+space), they have what appears to be a more general definition:

In functional analysis, a Schwartz space is a locally convex topological vector space $E$ with the property that whenever $U$ is an absolutely convex neighbourhood of $0$ then it contains another, say $V$, such that $U$ maps to a precompact set in the normed vector space $E_V$.

For the record, I'm an expert on neither of these spaces; however, my naive intuition seems to suggest that the nLab space is somehow "less concrete" than the typical definition involving rapidly-decreasing derivatives. I am unable to seek out more understanding on this, however, because no reference is given at nLab.

So are these notions of Schwartz spaces related? Are they the same? Is there any literature that mentions the nLab definition?

Any insight would be much appreciated.

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2 Answers 2

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Disclaimer: I am not, and never have claimed to be, a functional analyst. However, I used functional analysis a lot in my work in differential topology and geometry. In particular, I read and re-read the mighty tome The Convenient Setting of Global Analysis extensively.

That book contains an appendix with a summary of all the functional analysis used. In particular, on page 585 (online edition), after Result 52.24, appears the definition:

A locally convex space $E$ is called Schwartz if each absolutely convex neighborhood $U$ of $0$ in $E$ contains another one $V$ such that the induced mapping $E_{ ( U )} \to E_{ ( V )}$ maps $U$ into a precompact set.

and this is the definition that appears on the nLab (unsurprisingly, given who is the original author of that page).

(Note: I think that the map $E_{(U)} \to E_{(V)}$ is actually the wrong way around. Reading elsewhere in that appendix, $E_{(U)}$ is the normed vector space formed by taking the Minkowski functional associated to $U$ as a semi-norm and making it a norm by quotienting out by the null space. Therefore an inclusion $V \to U$ should induce a mapping $E_{(V)} \to E_{(U)}$ and one would examine the image of $V$ in $E_{(U)}$ and want that to be precompact. I therefore think that it should read:

A locally convex space $E$ is called Schwartz if each absolutely convex neighborhood $U$ of $0$ in $E$ contains another one $V$ such that the induced mapping $E_{ ( V )} \to E_{ ( U )}$ maps $V$ into a precompact set.

)

The definition in The Convenient Setting ... is not directly cited, but the preceding result is about Schwarz spaces and is cited as Jarchow, 1981, 10.4.3, p202, Horvath, 1966, p277. So I would look there for more information.

The purpose of this definition is to try to capture the functional analytical essence of the Schwartz space (of rapidly decreasing functions). Thus anything one can say about the Schwartz space (in Functional Analysis) can be said about a Schwartz space (and, probably, vice versa). This fits in with a general theme of Functional Analysis wherein one says "This space has all these nice properties, what is it about this space which makes it have them? What other spaces have these properties? This looks like a useful class of spaces, let's give them a funky name.".

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  • $\begingroup$ How does it relate to the fact the usual Schwartz space is the intersection of the Banach spaces $B_{k,m}$ with norm $\|\phi\|_{B_{k,m}} = \sum_{n \le k} \sup_x (1+|x|^m) |\phi^{(n)}(x)|$ ? $\endgroup$
    – reuns
    Commented Aug 14, 2017 at 20:52
  • $\begingroup$ @reuns In short, for each $(k,m)$ there will be an $(l,n)$ such that the natural map $B_{(l,n)} \to B_{(k,m)}$ maps the unit ball of $B_{(l,n)}$ into a precompact subset. $\endgroup$ Commented Aug 15, 2017 at 10:21
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A few years late to this party... :)

Still, it might be useful to note that the "more general condition" is approximately asserting that the TVS is a (projective) limit of Banach spaces with compact transition maps. Sometimes this is the definition of a "nuclear" space. EDIT: and, as @Jochen aptly notes, this is not a good definition of nuclear space. Sufficient compositions of the transition maps must be trace class. For me, the most intelligible nuclear spaces are projective limits of Hilbert spaces, with Hilbert-Schmidt transition maps... knowing that composition of two H-S maps is trace-class. I like this version better than the Banach space version, because Hilbert-Schmidt maps are easier to characterize, and characterizing trace-class maps (on Hilbert spaces) as compositions of two Hilbert-Schmidt is more appealing to me than having to quantify over bases. :)

Among other virtues of such spaces $V$, a Schwartz Kernel Theorem is valid: every continuous linear map from $V$ to the weak dual $W^*$ of another such space $W$ is given by an element of $V^*\otimes W^*$... and that tensor product exists, in a genuine categorical sense (not merely as projective or injective tensor product... this goes back to Grothendieck's early work in the early 1950's).

Just for contrast, (infinite-dimensional) Hilbert spaces have no genuine tensor product, essentially because not all maps between them are Hilbert-Schmidt! :) So, it's not a pathology! :) (Note: the popular "tensor product of Hilbert spaces" is not a fully categorically-correct tensor product!)

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  • $\begingroup$ No, for nuclear locally convex spaces you need even better transition maps, namely that they are nuclear operators between Banach spaces (then one can prove that they factorize through a Hilbert space which is convenient since the theory of operatos between Hilbert spaces is easier). There are Fréchet-Schwartz spaces which fail to be nuclear, for example Köthe sequence spaces. This can be found, e.g., in book Introduction to Functional Analysis of Meise and Vogt. However, the Schwartz space of rapidely decreasing smooth functions is nuclear. $\endgroup$
    – Jochen
    Commented Oct 15, 2023 at 13:53

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