This morning I was looking for the book “Vector measures” by Diestel and Uhl in our local math library. Although it was not there I found the book “Sequences and Series in Banach Spaces” by Joseph Diestel. This (very nicely written) book on somehow advanced theory of Banach spaces contains a number of facts of the type “yeah, that is true, although not easy to see; however, I do not know any reference for…”.

**1. The dual of **

The question on how the dual space of looks like seems to be of interest for many people (e.g. questions on it are frequently asked on the web, e.g. at math.stackexchange). Diestel’s book has the chapter “The Classical Banach Spaces” and in the section “The Classical Nonreflexive Sequence Spaces” he describes helpful properties of the space , and .

The spaces and have the property that maps into them can be extended to larger domains:

Theorem 1Let be a linear subspace of a Banach space and let be bounded and linear. Then there is an extension of having the same norm as .

*Proof:* Since is bounded, it holds for all that and hence the mappings are all elements in the space —we denote them by . This gives the representation

By Hahn-Banach there exist the extensions of onto (with equal norm) and hence, an extension of is given by

works.

For something similar holds but only for separable spaces (with a more involved proof):

Theorem 2Let be a linear subspace of a separable Banach space and let be bounded and linear. Then there is an extension of .

Then Diestel moves on to the dual space of : (which stands for the bounded and (finitely) additive measures). Although this topic is also treated in other classical book (as “Linear Operators” by Dunford and Schwartz), the exposition in Diestel’s book was the most lively I came across.

Start with an . For a subset of the natural numbers the characteristic function belongs to and hence, we can evaluate . Of course, is disjoint additive in and for disjoint we have

This shows that indeed contains *finitely additive measures*. I’d like to quote Diestel directly:

A scalar-valued measure being bounded and additive is very like a countably additive measure and is not [...] at all pathological.

Now we denote by the Banach space space of bounded additive scalar-valued measures on endowed is with the variational norm (which can be defined through the Hahn-Jordan decomposition and is, in a nutshell, the measure of the positive set plus the measure of the negative set)

For and disjoint finite subsets of the natural numbers it holds for every that

and hence

We see that adds up the measures of countably many disjoint sets, especially, is an absolutely convergent sequence. However, the sum does not have to be the right one: may be smaller than . Diestel says that it is not fair to blame the measures for this probable defect but it is “a failure on the part of the underlying field of sets”. He goes on the make this precise by the use of the Stone representation theorem which assigns to any Boolean algebra (in our case the algebra of subsets of ) the appropriately topologized set of ultrafilters and in this set he considers the Boolean algebra of the sets which are “clopen” (which is isomorphic to the original algebra ). He then considers not acting on but its identical twin on and shows that there one needs to work with the *closure* of the union of sets and this makes the identical twin of even countably additive. In this sense, the lack of countable additivity is not a failure of but of , so to say.

**2. Weak convergence in is strong **

As a unique feature of I’d like to quote Diestel again:

Face it: the norm of a vector in is as big as it can be () if respect for the triangle inequality and the “unit” vector is to be preserved.

The following theorems hold:

Theorem 3For every separable Banach space there exists a bounded and linear operator which is onto.

Theorem 4Let be a Banach space and bounded, linear and onto. Then contains a subspace that is isomorphic to and complemented.

Probably the most stunning fact about is that weak and strong convergence coincide. To show this Diestel used heavy machinery, namely Phillips’ Lemma.

Lemma 5 (Phillips’ Lemma)Let be in and satisfy for every . Then

Theorem 6 (Schur’s Theorem)In it holds that iff .

*Proof:* As always, strong convergence implies weak convergence and we only have to show the converse. Of course we considered and hence, by canonical embedding into the double dual, there is for every a and for every it holds that

Let be a weak null sequence in . Then it holds that

Now, Phillips’ Lemma gives

By the way: Does anyone has a reference for a more direct/elementary proof of Schur’s Theorem?

October 5, 2011 at 7:36 pm

Concerning your question at the end: I cannot give you a reference, but some time ago I’ve written an elementary proof of Schur’s Theorem for some (never finished) lecture notes. I can send it to you tomorrow.

Markus

October 5, 2011 at 8:12 pm

This would be great! If you don’t mind, I could post the proof here..