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 [1]). 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 [2].

Theorem(A.N. Kolmogorov) A tvs is normable if and only if it is separated and contains a bounded convex nbhd of zero.

*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

(1)

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.

Pingback: On the Normability of the Space of Measurable Functions | Math by Matt