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We denote by $\mathcal{S}(\mathbb{R})$ the space of smooth and rapidly decreasing functions. We define on $\mathcal{S}(\mathbb{R})$ the family of semi-norms $$\lVert \varphi \lVert_{n,m} = \lVert (1+|\cdot|^m) \varphi^{(n)} \lVert_\infty.$$ This family of semi-norms defines a topology $\tau$ on $\mathcal{S}(\mathbb{R})$. The space $(\mathcal{S}(\mathbb{R}),\tau)$ is then well-known to be nuclear.

If $\mathcal{M}(\mathbb{R})$ is the space of measurable functions in $\mathbb{R}$, I define the space $\mathcal{R}(\mathbb{R})$ of rapidly decreasing functions (without smoothness condition) by $$\mathcal{R}(\mathbb{R}) = \{ \varphi \in \mathcal{M}(\mathbb{R}), \ \lVert \varphi \lVert_{0,m} < \infty \}.$$ Moreover, the family of (semi-)norms $(\lVert \cdot \lVert_{0,m})_{m\in \mathbb{N}}$ defines a topology $\tau'$ on $\mathcal{R}(\mathbb{R})$.

Question: Is $(\mathcal{R}(\mathbb{R}),\tau')$ a nuclear space?

NB. The proofs I found for the nuclearity of $\mathcal{S}(\mathbb{R})$ are essentially using the facts that the space $s$ of sequences with quick decay is nuclear and $\mathcal{S}(\mathbb{R})$ is homeomorphic to $s$.

Several authors are claiming that the nuclear structure is strongly related with the condition of infinite smoothness. The space $\mathcal{D}(\mathbb{R})$ of compactly supported and smooth function (with the inductive topology coming from the spaces $\mathcal{D}([-n,n])$), and the space $\mathcal{C}^\infty(\mathbb{R})$ (projective limit of the spaces $\mathcal{D}([-n,n])$) are other famous examples of nuclear structure that go in that sense. This suggests that the space $\mathcal{R}(\mathbb{R})$ may be not nuclear or at least that the usual technics are not available.

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It might be of interest to know that spaces of the type you are interested in (intersections or unions of weighted $L^p$-spaces) were studied by Dieudonne and later authors. They are known as Köthe function spaces since they are the function space analogues of the Köthe sequence spaces (Stufenräume and gestufte Räume) and you can find recent articles on them via google. –  barcelos Apr 18 at 14:33
    
S(R) is homeomorphic to s itself, not just a subspace. See B. Simon "Distributions and their Hermite expansions" J. Math. Phys. 12 (1971), 140-148. –  Abdelmalek Abdesselam Apr 28 at 20:51

1 Answer 1

up vote 9 down vote accepted

Your space contains an isomorphic copy of $L^\infty([0,1])$ (consider the family of elements with support in the interval) and so is not nuclear.

By the way an explanation of the relationship between smoothness and nuclearity is that many spaces of test functions are generated in a natural way by differential operators as described in a short but important article by A. Pietsch (Math. Ann. vol. 164). This article contains a simple criterion for such spaces to be nuclear, based on the spectral properties of the operators as unbounded, self-adjoint ones on Hilbert space and displays some of the standard spaces as explicit examples.

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Thanks a lot. For the answer and the paper. –  Goulifet Feb 19 at 12:29
    
@Goulifet. Added as an afterthought. Pietsch' method also explains the ubiquity of the space $s$ in the theory of test functions. The spaces introduced in his article are isomorphic to this one whenever the eigenvalues of the differential operator are asymptotically like a (positive) power of $n$ (this is not in the quoted article but is an easy consequence of its methods). Of course, many of the classical partial differential operators of theoretical physics (Laplace and Schrödinger) satisfy this condition. –  alpha Feb 19 at 17:48
    
Another reason for the ubiquity is of course the Komura-Komura theorem saying that every nuclear locally convex space is isomorphic to a subspace of $s^I$ for some index set $I$. A good reference is the book Introduction to Functional Analysis of Meise and Vogt. –  Jochen Wengenroth Feb 20 at 7:55
    
What is the name of that paper by Pietsch? –  Matthias Ludewig Apr 29 at 10:53

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