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 ${\ell^\infty}$

The question on how the dual space of ${\ell^\infty}$ 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 ${c_0}$, ${\ell^1}$ and ${\ell^\infty}$.

The spaces ${c_0}$ and ${\ell^\infty}$ have the property that maps into them can be extended to larger domains:

Theorem 1 Let ${Y}$ be a linear subspace of a Banach space ${X}$ and let ${T:Y\rightarrow\ell^\infty}$ be bounded and linear. Then there is an extension ${S:X\rightarrow\ell^\infty}$ of ${T}$ having the same norm as ${T}$.

Proof: Since ${T}$ is bounded, it holds for all ${n\in{\mathbb N}}$ that ${|Ty|_n\leq \|Ty\|_\infty\leq c\|y\|_Y}$ and hence the mappings ${y\mapsto (Ty)_n}$ are all elements in the space ${Y^*}$—we denote them by ${y^*_n}$. This gives the representation

$\displaystyle Ty = (y^*_n(y))_n.$

By Hahn-Banach there exist the extensions ${x^*_n}$ of ${y^*_n}$ onto ${X}$ (with equal norm) and hence, an extension of ${T}$ is given by

$\displaystyle Sx = (x^*_n(x))_n$

works. $\Box$

For ${c_0}$ something similar holds but only for separable spaces ${X}$ (with a more involved proof):

Theorem 2 Let ${Y}$ be a linear subspace of a separable Banach space ${X}$ and let ${T:Y\rightarrow c_0}$ be bounded and linear. Then there is an extension ${S:X\rightarrow c_0}$ of ${T}$.

Then Diestel moves on to the dual space of ${\ell^\infty}$: ${ba(2^{\mathbb N})}$ (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 ${x^*\in (\ell^\infty)^*}$. For a subset ${\Delta}$ of the natural numbers the characteristic function ${\chi_\Delta}$ belongs to ${\ell^\infty}$ and hence, we can evaluate ${x^*(\chi_\Delta)}$. Of course, ${x^*(\chi_\Delta)}$ is disjoint additive in ${\Delta}$ and for disjoint ${\Delta_1,\dots,\Delta_n}$ we have

$\displaystyle \begin{array}{rcl} \sum_{i=1}^n |x^*(\chi_{\Delta_i})| &= &\sum_{i=1}^n x^*(\chi_{\Delta_i})\, \text{sgn}(x^*(\chi_{\Delta_i}))\\ & = &x^*(\sum_{i=1}^n \text{sgn}(x^*(\chi_{\Delta_i}))\,\chi_{\Delta_i})\\ & \leq &\|x^*\|\,\|\sum_{i=1}^n \text{sgn}(x^*(\chi_{\Delta_i}))\,\chi_{\Delta_i}\|_\infty\leq\|x^*\|. \end{array}$

This shows that ${(\ell^\infty)^*}$ 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 ${ba(2^{\mathbb N})}$ the Banach space space of bounded additive scalar-valued measures on ${{\mathbb N}}$ endowed is with the variational norm ${\|\cdot\|_{ba}}$ (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 ${\mu\in ba(2^{\mathbb N})}$ and disjoint finite subsets ${\Delta_k}$ of the natural numbers it holds for every ${n}$ that

$\displaystyle \sum_{i=1}^n |\mu(\Delta_i)|\leq \|\mu\|_{ba}$

and hence

$\displaystyle \sum_{i=1}^\infty |\mu(\Delta_i)|\leq \|\mu\|_{ba}.$

We see that ${\mu}$ adds up the measures of countably many disjoint sets, especially, ${\sum_{i=1}^\infty \mu(\Delta_i)}$ is an absolutely convergent sequence. However, the sum does not have to be the right one: ${\sum_{i=1}^\infty \mu(\Delta_i)}$ may be smaller than ${\mu(\bigcup_{i=1}^\infty \Delta_i)}$. Diestel says that it is not fair to blame the measures ${\mu}$ 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 ${\Sigma=2^{\mathbb N}}$ of subsets of ${{\mathbb N}}$) the appropriately topologized set of ultrafilters and in this set he considers the Boolean algebra ${\mathcal{S}(\Sigma)}$ of the sets which are “clopen” (which is isomorphic to the original algebra ${\Sigma}$). He then considers ${\mu}$ not acting on ${\Sigma}$ but its identical twin ${\hat\mu}$ on ${\mathcal{S}(\Sigma)}$ and shows that there one needs to work with the closure of the union of sets and this makes the identical twin of ${\mu}$ even countably additive. In this sense, the lack of countable additivity is not a failure of ${\mu}$ but of ${\Sigma}$, so to say.

2. Weak convergence in ${\ell^1}$ is strong

As a unique feature of ${\ell^1}$ I’d like to quote Diestel again:

Face it: the norm of a vector ${\sum_n t_n e_n}$ in ${\ell^1}$ is as big as it can be (${\|\sum_n t_n e_n\|_1 = \sum_n |t_n|}$) if respect for the triangle inequality and the “unit” vector is to be preserved.

The following theorems hold:

Theorem 3 For every separable Banach space ${X}$ there exists a bounded and linear operator ${T:X\rightarrow\ell^1}$ which is onto.

Theorem 4 Let ${X}$ be a Banach space and ${T:X\rightarrow\ell^1}$ bounded, linear and onto. Then ${X}$ contains a subspace that is isomorphic to ${\ell^1}$ and complemented.

Probably the most stunning fact about ${\ell^1}$ is that weak and strong convergence coincide. To show this Diestel used heavy machinery, namely Phillips’ Lemma.

Lemma 5 (Phillips’ Lemma) Let ${\mu_n}$ be in ${ba(2^{\mathbb N})}$ and satisfy ${\lim_{n\rightarrow\infty}\mu_n(\Delta)=0}$ for every ${\Delta\subset{\mathbb N}}$. Then

$\displaystyle \lim_n \sum_k |\mu_n(\{k\})|=0.$

Theorem 6 (Schur’s Theorem) In ${\ell^1}$ it holds that ${x_n\rightharpoonup x}$ iff ${x_n\rightarrow x}$.

Proof: As always, strong convergence implies weak convergence and we only have to show the converse. Of course we considered ${\ell^1=\ell^1({\mathbb N})}$ and hence, by canonical embedding into the double dual, there is for every ${x\in\ell^1}$ a ${\mu_x\in ba(2^{\mathbb N})}$ and for every ${\Delta\subset {\mathbb N}}$ it holds that

$\displaystyle \mu_x(\Delta) = \sum_{k\in\Delta} x(k).$

Let ${(x_n)}$ be a weak null sequence in ${\ell^1}$. Then it holds that

$\displaystyle \begin{array}{rcl} \lim_{n\rightarrow\infty} \mu_{x_n}(\Delta) &=&\lim_{n\rightarrow\infty}\sum_{k\in\Delta}x_n(k)\\ & = & \lim_{n\rightarrow\infty}\langle \chi_\Delta,x_n\rangle_{\ell^\infty\times\ell^1} = 0 \end{array}$

Now, Phillips’ Lemma gives

$\displaystyle 0 = \lim_n \sum_k |\mu_n(\{k\})|= \lim_n \sum_k |x_n(k))|=\lim_n \|x_n\|_1.$

$\Box$

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