## Topology on the space of Schwartz Distributions

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If we equip the Schwartz space $\mathcal{S}$ with its usual Fréchet space topology, then the space of continuous linear functionals $\mathcal{S}^\ast$ is known as the space of Schwartz distributions or tempered distributions. If we equip this space with the strong topology, is there anything we can say about the resulting topological vector space? Evidently, the resulting space will not be a Fréchet space, but perhaps it will have other nice properties. In particular, I am interested in the space of continous linear operators on $\mathcal{S}^\ast$. Is there anything interesting we can say about this space?

Unfortunately, a quick google search did not turn up many sources that dealt with the particulars of the topology on $\mathcal{S}^\ast$, much less the topology on the space of continuous linear operators on $\mathcal{S}^\ast$, so a point in the right direction to a reference would also be greatly appreciated.

EDIT: After thinking about this more deeply, I realize that I am interested in a specific type of operator on $\mathcal{S}^\ast$. $\mathcal{S}$ occurs naturally inside of $L^2$, so after identifying the dual of $L^2$ with itself via the Riesz Representation Theorem, we can in turn regard $\mathcal{S}$ as a subspace of $\mathcal{S}^\ast$. With this in mind, I am interested in the operators on $\mathcal{S}^\ast$ that restrict to operators on $\mathcal{S}$.

The motivation for this question comes from quantum mechanics, where I have in mind the position and momentum operators acting on $\mathcal{S}^\ast$. I am thus interested in the operator algebra they generate. Furthermore, these of course restrict to operators on $\mathcal{S}$, and so I am likewise interseted in the operator algebra of operators on $\mathcal{S}^\ast$ that restrict to operators on $\mathcal{S}$. In particular, I would like to abstractly characterize this space.

As this is the natural space for observables in quantum mechanics (whether physicsts realize it or not), there has to be at least something known about this space. . .

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The strong dual of a Frechet space is what is called a DF-space: see e.g. ncatlab.org/nlab/show/DF+space I'm afraid I don't know about the space of all linear operators on DF-spaces, but hopefully someone will come along who knows more about this – Yemon Choi Nov 4 2011 at 5:36
Mathematical physicists are well aware of the relevance of this situaton to quantum mechanics.  I recommend you google "rigged Hilbert space" and "Gelfand triple".   What you are looking at is an (probably the) example of such structures. Note that it arises from the following data.  A Hilbert space and an unbounded operator thereon (the standard one-dimensional Schrödinger operator).  Any such operator leads to corresponding structures.  The fact that the above operator has discrete spectrum and that the eigenvalues grow like a power of $n$ (in this case the square) makes life simpler. – jbc Nov 28 at 7:54

## 5 Answers

A detailed study of topological properties of the space $S^*$ and other similar spaces of distributions is given in the book

I. M. Gel'fand and G. E. Shilov, Generalized functions. Vol. 2: Spaces of fundamental and generalized functions, Academic Press, 1968.

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Perhaps, the following will be interesting for you: the strong topology on S*, obviously, coincides with the topology of uniform convergence on totally bounded sets, and this means that S*, being endowed with this topology, is what is called stereotype space. :) The definition is as follows. For a locally convex space X let us denote by X* the dual space (of functionals) endowed with the topology of uniform convergence on totally bounded sets. Then X is said to be {\it stereotype}, if X** is naturally isomorphic (as a locally convex space) to X.

By the way, this kind of duality allows to consider linear continuous operators X*$\to$X* as linear continuous operators X$\to$X. :)

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Could you explain in more detail how we can consider linear operators on $X^\ast$ as linear opeerators on $X$?  I am thinking of the more familiar example of $L^p$ and $L^q$ for $1<p<q<\infty$ Hölder conjugates, and in general I don't think linear operators on $L^p$ correspond to operators on $L^q$, at least not in any way I know of.  Other the other hand, I see how you can obtain a sort-of adjoint operator on $X$ given an operator on $X^\ast$.  Was this what you had in mind? – Jonathan Gleason Nov 4 2011 at 21:36
Suppose X is stereotype, and X* is its dual space of linear continuous functionals on X (again with the topology of uniform convergence on totally bounded sets on X). Then to each operator A:X$\to$X one can assign a dual operator A*:X*$\to$X* by formula A*f(x)=f(Ax), x$\in$X, f$\in$X*. The properties of A* are totally determined by the properties of A. On the other hand, since X is stereotype, the operation of taking dual operator A$\mapsto$A* is bijective, and the properties of A are totally determined by properties of A*. – Sergei Akbarov Nov 5 2011 at 8:35
Actually, this is a usual trick in all the theories of reflexivity (I think I should have told about this from the very beginning). For example, the spaces $L^p$, which you mentioned, are (not only stereotype but also) reflexive in the usual sence (for $1<p<\infty$), like reflexive locally convex spaces. And the Schwartz space S is also reflexive in this traditional sence -- so, perhaps, there were no need to begin a talk about stereotype theory.  – Sergei Akbarov Nov 5 2011 at 11:06
But the advantage of stereotype theory is that it makes all the reasonable spaces in analysis reflexive (in new sence). In particular, all Frechet spaces are stereotype. – Sergei Akbarov Nov 5 2011 at 11:06

As regards your first question, the space of tempered distributions is indeed an example of a special class of locally convex spaces which, while not metrisable, have very nice properties. These are the Silva spaces which were investigated by the portuguese mathematician Sebastiao e Silva---they are, by definition, inductive limits of sequences of Banach spaces with compact interconnecting mappings---in your case, these are even nuclear. The original articles are rather inaccessible but Koethe's monograph on topological vector spaces has a chapter on these spaces. Since everything in sight in your application is nuclear, the operator spaces you are interested in can be represented as tensor products (in any of the standard tensor product topologies---in this case they all coincide). These facts were investigated in considerable detail by some of the great masters of functional analysis, for example, Koethe, Schwartz and Grothendieck and there is a wealth of material on this topic in their works---for example, Koethe's treatise mentioned above, Schwartz' sequel to his classical "Th\'eorie des Distributions" (in which he considers vector-valued distributions---this is easily available online) and Grothendick's thesis. One can find more accesssible representations in the recent secondary literature on locally convex spaces.

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Just as a variation on the other answers: the Schwartz space can easily be written as a (projective) limit of Hilbert spaces $V_s$, basically requiring that both a function and its Fourier transform be in a Levi-Sobolev space. The inclusions for $s>t$ are compact (in fact, trace-class, seen by proving that they're composites of Hilbert-Schmidt) inclusions. This adds a bit to the assertion that it is Frechet, which would indeed give a proj lim of Banach spaces, but if we can have Hilbert spaces, it's even better. Then, as in the question, identifying $L^2(\mathbb R)$ with its own dual (up to complex conjugation, anyway), the dual of $V_s$ for positive $s$ is $V_{-s}$... and we can give it the strong (=Hilbert-space) topology. It is not completely formal-categorical, but easy enough, to show that the dual of the limit is the colimit of the duals, however we topologize them. With the Hilbert-space topologies we obtain (yet another presentation of) the strong topology on tempered distributions.

It is completely formal that the dual of a colimit is the limit of duals, and a virtue of this set-up, with Hilbert spaces, is that the reflexivity is trivial.

Also, operators on $V_{-\infty}=\bigcup_s V_s$ are easy to understand in this presentation. The operators that stabilize $V_{+\infty}=\bigcap_s V_s$ must stabilize each Hilbert space $V_s$, etc.

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Concerning this: "I am likewise interseted in the operator algebra of operators on S* that restrict to operators on S. In particular, I would like to abstractly characterize this space"

I am not sure that I understand you correctly, but if you want to characterize the operators which are extensions from S to S*, then the following can be the answer. First, let us use some notations: for any x,y$\in$S we set $(x,y)=\int x(t)y(t)dt$, for any operator (everywhere operator will be linear and continuous) $A:S\to S$ we define a transposed operator $A^T:S\to S$ by formula $(Ax,y)=(x,A^Ty)$ (it does not always exist), the dual operator A*:S*$\to$ S* by formula A*f(x)=f(Ax) (it always exists) and for any operator B:S*$\to$S* its dual B*:S$\to$S by formula B*x(f)=Bf(x) (it also always exists). Let us also consider an operator F:S$\to$S*, Fx(y)=(x,y).

Then

1) B:S*$\to$S* is an extension of $A:S\to S$ iff $B\circ F=F\circ A$;

2) an operator A:S$\to$S can be extended to some operator B:S*$\to$S* iff there exists a transposed operator $A^T:S\to S$; in this case B=A$^T$*;

3) an operator B:S*$\to$S* is an extension of some operator A:S$\to$S iff for B* there exists a transposed operator B*$^T:S\to S$; in this case A=B*$^T$.

Well, the idea "abstractly characterize" is not a precisely defined, but I'll try to give you an idea of what I meant by way of example. If you take a subalgebra of bounded operators on a Hilbert space, this is a unital $C^*$-algebra. Similarly, the Gelfand-Naimark Theorem says that any abstract unital $C^*$-algebra is isomorphic to a subalgebra of all the bounded operators on some Hilbert space. In this way, the structure of a unital $C^*$-algebra abstractly characterizes bounded observables on a Hilbert space. Can something similar be done in this case? – Jonathan Gleason Nov 6 2011 at 0:12
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