This past week I have been refreshing my knowledge of topological vector spaces (tvs) while reading some papers on generalizations of the Mazur-Ulam theorem to metrizable tvs. I intend to upload a typed set of notes on the subject which I have been putting together, but they are still mostly handwritten. For now, I want to present a theorem of A.N. Kolmogorov, the Kolmogorov normability criterion, that gives necessary and sufficient conditions for a tvs to be normable (i.e. there exists a norm such that the norm topology coincides with ). Two reasons motivate me to share it: 1) the proof is elegant; 2) I have spent a fair amount of time this summer studying when topological spaces, in particular function spaces, are normable.
To minimize any confusion over terminology, I include the following definitions. A notational note: we denote the line segment between two points by .
Let be a vector space. A subset is…
- convex if for any .
- absorbing (or absorbent or radial) if for all , there exists a scalar such that .
- circled (or balanced if for any .
- bounded if for every neighborhood there exists a scalar such that .
It is a well-known that a Hausdorff tvs is metrizable if and only if it has a countable base of neighborhoods at zero (Theorem 5.12 ). However, there are plenty of familiar spaces which are metrizable, but normable. Consider the vector space of real sequences with addition and scalar multiplication defined pointwise. Define a metric by
The reader may verify that that is a complete, metrizable tvs. Let . Observe that as . Since a norm is positively homogeneous, as . But then for any , there does not exists an such that
Indeed, if be a standard basis vector such that , then for all , but the norm of the norm tends to as .
A more interesting example is the space of Lebesgue measurable functions with , where , lacks a norm equivalent to its natural topology. It turns out that the non-normability is intimately connected to the lack of nontrivial convex open sets in this space, a result known as Day’s theorem. The above example might suggest to us that convexity plays a role in any criterion for normability of a tvs.
Recall that a tvs is said to be separated if for any , there exists a nbhd of which does contain . Equivalently, is separated if there exists a nbhd of which does not contain . The following theorem of A.N. Kolmogorov characterizes the normability of a separated (not a priori Hausdorff) tvs in terms of the existence of a bounded convex neighborhood of zero. The proof presented here is that of .
Proof: The direction is obvious. For , suppose that is separated and is a bounded convex nbhd of zero. Taking a balanced nbhd of zero and noting that , we may assume that is balanced, bounded, and convex.
For , define a set of scalars ; define . I claim that contains nonzero scalars when . Indeed, for , there exists an open nbhd of zero such that . Since is bounded, there exists such that . As , . Although we know that contains nontrivial scalars, it is not obvious that contains an interval or disk when . We will establish this result below in order to define a norm structure on .
Define a nonnegative (possibly) extended-valued function . We will show that is a norm. It is clear that and the preceding claim shows that . I claim that
If , then the LHS is empty. If , then there exists a sequence of scalars increasing to . For , we have , as is balanced. Hence if , then . As a consequence, we obtain for all . Indeed, note that is a nbhd of zero, therefore absorbing. So given , there exists such that . It follows that .
For homogeneity, let be a nonzero scalar. Observe that by (1), if and only if . It follows that . If , then (1) implies . Letting , we obtain the reverse inequality.
So far we have not used the hypothesis that is convex, but we will need it to show that satisfies the triangle inequality. Fix . If either or , then the triangle inequality is trivial, so assume otherwise. Take sufficiently small so that . Then and . Since is convex,
whence . As , it follows that .
Lastly, we need to show that norm topology coincides with . Let be a ball of radius centered at zero. If , then , whence is -open. Conversely, suppose is an open nbhd of zero. Since is bounded, there is such that . If , then , whence .
1. S.K. Berberian, Lectures in Functional Analysis and Operator Theory, Spring Verlag 1974.