Topological Vector Wackness

As I was reading MathOverflow the other day I came across a fairly startling post: Are there non-reflexive vector spaces isomorphic to their bi-dual?

For those of you who aren’t familiar with the notation, in the basic duality theory of Banach spaces one begins with a space , then proceeds to consider the dual space of continuous linear ``functionals’’ on (e.g. maps from to the underlying scalar field). One immediately confronts the natural problem of topologizing in such a manner as to make it a Banach space of its own – and this is initially resolved by introducing the operator norm (where the supremum is over nonzero ). This construction is entirely general, and so can be repeated ad infinitum to get the Banach spaces , ,  etc.

In the special case when is a Hilbert space, we have a particularly strong duality theory. The Riesz representation says essentially that the map is a (conjugate-linear) isometric isomorphism , and it follows then that the map is an isometric isomorphism . In English: continous linear functionals of are simply evaluating elements of at points of .

In the more general theory of Banach spaces, one hopes for the case when the evaluation map gives an isometric isomorphism , and if this is the case we say that the space is ``reflexive’’. However, unlike in the simpler Hilbert space case, there are non-reflexive Banach spaces. Consider , whose dual is isomorphic to . We have that is separable but isn’t (just consider the indicator functions of subsets of – in the sup norm, distinct subsets give elements with of distance 1 from eachother). This is the crux of the matter, because it can be shown (via the Hahn-Banach theorem) that if a space is non-separable then its dual is non-separable. So is not isomorphic to .

But what’s really remarkable (and frightening), is that there are spaces which, although not isomorphic to their double dual via the natural ``evaluation’’ map, still can perhaps be isomorphic to their double dual! This is the fact mentioned in the MathOverflow post. How about them apples!

As if that isn’t frightening enough, there are in fact many naturally occuring topological vector spaces that are not even normed spaces.

The general theory of topological vector spaces considers a vector space with a topological structure (typically Hausdorff), such that addition and scalar multiplication are continuous operations. Perhaps one level of generality above normed vector spaces (or even pre-normed spaces, e.g. in which one can receive a norm if one mods out by a suitable subset) are the Frechet spaces, which (although definitions vary somewhat in the literature) are those topological vector spaces that are locally convex and in which the topology is induced by a complete, translation-invariant metric. This is a significantly weaker theory than that of Banach spaces, but broad enough to cover most of what occurs in practice.

In an undergraduate analysis course one is taught that the continuous functions on a compact set form a Banach space under the supremum norm. But what about continuous functions on an open set? Or perhaps the holomorphic functions on a region in the complex plane? Or the maps on some region in euclidean space? In all of these cases, there are reasonably behaved yet unbounded functions, and ``uniform convergence’’ (or something similar) on the whole domain would be a completely unreasonable requirement. The solution is to consider the behavior on compact sets – in each case, we have a reasonable theory to deal with compact domains, and we are able to approximate whatever open set we are defining our functions on by an interior sequence of compact sets.

As an example, in the case of continuous functions on an open set , we first find a sequence of compact sets such that and . One introduces the family of seminorms , and from these you can define a metric . It’s easy to see that is a Cauchy sequence for this metric iff as . That is, Cauchy sequences are those that converge uniformly on compacts. An entirely similar construction works in topologizing the holomorphic functions on a region, and if one modifies the seminorms in such a way as to encode information about the higher derivatives of a function, one can easily come up with other reasonable metrics on functions on some open set.

Now, one might ask whether or not there is some norm generating the topology of these spaces. It’s in fact the case that there is no hidden norm on these spaces, due to various fundamental topological obstructions. In the case of , for example, the space is not ``locally bounded’’ (in a certain non-metric sense), which essentially means that we cannot pick any reasonable unit ball to give us a norm.

As bad as this sounds, the rabbit hole goes much deeper. I mentioned that the Frechet spaces are the ``next level’’ of generality beyond the normed linear spaces. A tremendous part of the duality theory of topological vector spaces relies on the Hahn-Banach theorem and other consequences of convexity. However, there are seemingly natural spaces out there that are not locally convex! For example, the spaces for . We’ll consider the case where the underlying measure space is with Lebesgue measure. The most reasonable candidate for a metric on the space is (the lack of a normalizing exponent is so that we can at least have the triangle inequality). And indeed, this can be easily verified to be a metric.

Suppose is an open, nonempty convex neighborhood of 0 in . Our goal is to show that , which we can acheive by writing as the convex combination of elements in . Because our space is metric, contains some ball of radius , and for , we have for some (this is the moment when the crucial property of is being used). We may find a partition of such that (simply because indefinite integrals are continuous), and then define . We clearly see that and hence . By construction, ), and hence . Since this works for arbitrary , we see that .

Because convex sets pull back to convex sets via linear maps, there is a startling consequence of this lack of nontrivial convex sets in : there are no nontrivial continuous linear functionals on for ! That is, the only continuos linear map is the zero map. We see instantly that without local convexity that a reasonable duality theory is not only out of our reach, it’s in fact trivial.

As an aside, there is a certain amount of interest in functional spaces that aren’t locally convex. Consider the case common in probability, when one has random variables (or more generally, random elements of some separable metric space) and the commonly used notion of convergence in probability. It’s not too hard to come up with a metric for convergence in probability, but in many cases (certainly the standard ones when you are dealing with an atomless probability space) the space of random variables with the topology of convergence in probability is not locally convex. So one hits the same problem as in the case: there are significant issues with approaching the study of such spaces via traditional routes, since e.g. the classical notions of duality are just irrelevent. While the technical details are not particularly important to the classical theory of probability (in which we are primarily relying on various analytic tools, such as Chebyshev’s inequality), the more general theory of convergence in probability can arise when one considers applications to mathematical finance. I’m still a bit fuzzy on this, but at the core of the mathematical theory of asset pricing is the change of measure to risk-neutral probability, so that martingale methods may be applied. Changing measure does not preserve notions such as or almost sure convergence, but for measures with the same null sets the transformation does preserve convergence in probability. In this context, convergence in probability is the most apparently reasonable notion of convergence.