# The Dunford-Pettis theorem

## 1 Weak topology and weak-* topology

If $(E,\tau )$ is a topological vector space, we denote by ${E}^{*}$ the set of continuous linear maps $E\to \u2102$, the dual space of $E$. The weak topology on $E$, denoted $\sigma (E,{E}^{*})$, is the coarsest topology on $E$ with which each function $x\mapsto \lambda x$, $\lambda \in {E}^{*}$, is continuous $E\to \u2102$. Thus, $\sigma (E,{E}^{*})\subset \tau $. If $(E,\tau )$ is a locally convex space, it follows by the Hahn-Banach separation theorem that ${E}^{*}$ separates $X$, and hence $|\lambda |,\lambda \in {E}^{*}$, is a separating family of seminorms on $E$ that induce the topology $\sigma (E,{E}^{*})$. Therefore, if $(E,\tau )$ is a locally convex space, then $(E,\sigma (E,{E}^{*}))$ is a locally convex space.

If $(E,\tau )$ is a topological vector space, the weak-* topology on ${E}^{\mathrm{*}}$, denoted $\sigma ({E}^{*},E)$, is the coarsest topology on ${E}^{*}$ with which each function $\lambda \mapsto \lambda x$, $x\in E$, is continuous ${E}^{*}\to \u2102$. It is a fact that ${E}^{*}$ with the topology $\sigma ({E}^{*},E)$ is a locally convex space.

If $E$ is a normed space, then ${\parallel \lambda \parallel}_{op}={sup}_{\parallel x\parallel \le 1}|\lambda x|$ is a norm on the dual space ${E}^{*}$, and that ${E}^{*}$ with this norm is a Banach space. The Banach-Alaoglu theorem states that $\{\lambda \in {E}^{*}:{\parallel \lambda \parallel}_{op}\le 1\}$ is a compact subset of $({E}^{*},\sigma ({E}^{*},E))$.

If $(X,\mathrm{\Sigma},\mu )$ is a $\sigma $-finite measure space, for
$g\in {L}^{\mathrm{\infty}}(\mu )$ define ${\varphi}_{g}\in {({L}^{1}(\mu ))}^{*}$ by
${\varphi}_{g}(f)={\int}_{X}fg\mathit{d}\mu $. The map
$g\mapsto {\varphi}_{g}$ is an isometric isomorphism
${L}^{\mathrm{\infty}}(\mu )\to {({L}^{1}(\mu ))}^{*}$.^{1}^{1}
1
Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications,
second ed., p. 190, Theorem 6.15.

Let $(X,\mathrm{\Sigma},\mu )$ be a probability space. If $\mathrm{\Psi}\in {({L}^{\mathrm{\infty}}(\mu ))}^{*}$ and $A\mapsto \mathrm{\Psi}({\chi}_{A})$ is countably additive on $\mathrm{\Sigma}$, then there is some $f\in {L}^{1}(\mu )$ such that

$$\mathrm{\Psi}(g)={\int}_{X}gf\mathit{d}\mu ,g\in {L}^{\mathrm{\infty}}(\mu ),$$ |

and ${\parallel \mathrm{\Psi}\parallel}_{op}={\parallel f\parallel}_{1}$.^{2}^{2}
2
V. I. Bogachev, Measure Theory, volume I,
p. 263, Proposition 4.2.2.
Also, an additive function $F$ on an algebra of sets $\mathcal{A}$ is countably additive
if and only if whenever ${A}_{n}$ is a decreasing sequence of elements of $\mathcal{A}$ with
${\bigcap}_{n=1}^{\mathrm{\infty}}{A}_{n}=\mathrm{\varnothing}$, we have ${lim}_{n\to \mathrm{\infty}}F({A}_{n})=0$.^{3}^{3}
3
V. I. Bogachev, Measure Theory, volume I,
p. 9, Proposition 1.3.3.
Using that $\mu $ is countably additive we get the following.

###### Theorem 1.

Suppose that $(X,\mathrm{\Sigma},\mu )$ be a probability space and that $\mathrm{\Psi}\in {({L}^{\mathrm{\infty}}(\mu ))}^{*}$, and suppose that for each $\u03f5>0$ there is some $\delta >0$ such that $E\in \mathrm{\Sigma}$ and $\mu (E)\le \delta $ imply that $|\mathrm{\Psi}({\chi}_{A})|\le \u03f5$. Then there is some $f\in {L}^{1}(\mu )$ such that

$$\mathrm{\Psi}(g)={\int}_{X}gf\mathit{d}\mu ,g\in {L}^{\mathrm{\infty}}(\mu ).$$ |

## 2 Normed spaces

If $E$ is a normed space, its dual space ${E}^{*}$ with the operator norm is a Banach space, and ${E}^{**}={({X}^{*})}^{*}$ with the operator norm is a Banach space. Define $i:E\to {E}^{**}$ by

$$i(x)(\lambda )=\lambda (x),x\in E,\lambda \in {E}^{*}.$$ |

It follows from the Hahn-Banach extension theorem that $i:E\to {E}^{**}$ is an isometric linear map.

If $E$ and $F$ are normed spaces and $T:E\to F$ is a bounded linear map, we define the transpose ${T}^{*}:{F}^{*}\to {E}^{*}$ by ${T}^{*}\lambda =\lambda \circ T$ for $\lambda \in {F}^{*}$. If $T$ is an isometric isomorphism, then ${T}^{*}:{F}^{*}\to {E}^{*}$ is an isometric isomorphism, where ${E}^{*}$ and ${F}^{*}$ are each Banach spaces with the operator norm. In particular, we have said that when $(X,\mathrm{\Sigma},\mu )$ is a $\sigma $-finite measure space, then the map $\varphi :{L}^{\mathrm{\infty}}(\mu )\to {({L}^{1}(\mu ))}^{*}$ defined for $g\in {L}^{\mathrm{\infty}}(\mu )$ by

$${\varphi}_{g}(f)={\int}_{X}fg\mathit{d}\mu ,f\in {L}^{1}(\mu ),$$ |

is an isometric isomorphism, and hence ${\varphi}^{*}:{({L}^{1}(\mu ))}^{**}\to {({L}^{\mathrm{\infty}}(\mu ))}^{*}$ is an isometric isomorphism. Therefore, for $E={L}^{1}(\mu )$ we have that

$${\varphi}^{*}\circ i:{L}^{1}(\mu )\to {({L}^{\mathrm{\infty}}(\mu ))}^{*}$$ | (1) |

is an isometric linear map. For $f\in {L}^{1}(\mu )$ and $g\in {L}^{\mathrm{\infty}}(\mu )$,

$({\varphi}^{*}\circ i)(f)(g)$ | $=({\varphi}^{*}(i(f)))(g)$ | ||

$=(i(f)\circ \varphi )(g)$ | |||

$=i(f)({\varphi}_{g})$ | |||

$={\varphi}_{g}(f).$ |

The Eberlein-Smulian theorem states that if $E$ is a normed space and $A$ is a subset of $E$, then $A$ is weakly
compact if and only if $A$ is weakly sequentially compact.^{4}^{4}
4
Robert E. Megginson, An Introduction to Banach
Space Theory, p. 248, Theorem 2.8.6.

## 3 Equi-integrability

Let $(X,\mathrm{\Sigma},\mu )$ be a probability space and let $\mathcal{F}$ be a subset of ${L}^{1}(\mu )$. We say that $\mathcal{F}$ is equi-integrable if for every $\u03f5>0$ there is some $\delta >0$ such that for any $A\in \mathrm{\Sigma}$ with $\mu (A)\le \delta $ and for all $f\in \mathcal{F}$,

$${\int}_{A}|f|\mathit{d}\mu \le \u03f5.$$ |

If $\mathcal{F}$ is a bounded subset of ${L}^{1}(\mu )$, it is a fact that $\mathcal{F}$ being equi-integrable is equivalent to

$$\underset{C\to \mathrm{\infty}}{lim}\underset{f\in \mathcal{F}}{sup}{\int}_{\{|f|>C\}}|f|\mathit{d}\mu =0.$$ | (2) |

The following theorem gives a condition under which a sequence of integrable functions is bounded and
equi-integrable.^{5}^{5}
5
V. I. Bogachev, Measure Theory, volume I, p. 269, Theorem 4.5.6.

###### Theorem 2.

Let $(X,\mathrm{\Sigma},\mu )$ be a probability space and let ${f}_{n}$ be a sequence in ${L}^{1}(\mu )$. If for each $A\in \mathrm{\Sigma}$ the sequence ${\int}_{A}{f}_{n}\mathit{d}\mu $ has a finite limit, then $\{{f}_{n}\}$ is bounded in ${L}^{1}(\mu )$ and is equi-integrable.

## 4 The Dunford-Pettis theorem

A subset $A$ of a topological space $X$ is said to be relatively compact if $A$ is contained in some compact subset of $X$. When $X$ is a Hausdorff space, this is equivalent to the closure of $A$ being a compact subset of $X$.

The following is the Dunford-Pettis theorem.^{6}^{6}
6
V. I. Bogachev, Measure Theory, volume I,
p. 285, Theorem 4.7.18; Fernando Albiac and
Nigel J. Kalton, Topics in Banach Space Theory, p. 109, Theorem 5.2.9;
R. E. Edwards, Functional Analysis: Theory and Applications, p. 274, Theorem 4.21.2;
P. Wojtaszczyk, Banach Spaces for Analysts,
p. 137, Theorem 12; Joseph Diestel, Sequences and Series in Banach Spaces, p. 93; François Trèves, Topological Vector Spaces, Distributions and Kernels, p. 471, Theorem 46.1.

###### Theorem 3 (Dunford-Pettis theorem).

Suppose that $(X,\mathrm{\Sigma},\mu )$ is a probability space and that $\mathcal{F}$ is a bounded subset of ${L}^{1}(\mu )$. $\mathcal{F}$ is equi-integrable if and only if $\mathcal{F}$ is a relatively compact subset of ${L}^{1}(\mu )$ with the weak topology.

###### Proof.

Suppose that $\mathcal{F}$ is equi-integrable, and let $T={\varphi}^{*}\circ i:{L}^{1}(\mu )\to {({L}^{\mathrm{\infty}}(\mu ))}^{*}$ be the isometric linear map in (1), for which

$$T(f)(g)={\int}_{X}fg\mathit{d}\mu ,f\in {L}^{1}(\mu ),g\in {L}^{\mathrm{\infty}}(\mu ).$$ |

Then $T(\mathcal{F})$ is a bounded subset of ${({L}^{\mathrm{\infty}}(\mu ))}^{*}$, so is contained in some closed ball $B$ in ${({L}^{\mathrm{\infty}}(\mu ))}^{*}$. By the Banach-Alaoglu theorem, $B$ is weak-* compact, and therefore the weak-* closure $\mathscr{H}$ of $T(\mathcal{F})$ is weak-* compact. Let $F\in \mathscr{H}$. There is a net ${F}_{\alpha}=T({f}_{\alpha})$ in $T(\mathcal{F})$, $\alpha \in I$, such that for each $g\in {L}^{\mathrm{\infty}}(\mu )$, ${F}_{\alpha}(g)\to F(g)$, i.e.,

$${\int}_{X}{f}_{\alpha}g\mathit{d}\mu \to F(g),g\in {L}^{\mathrm{\infty}}(\mu ).$$ | (3) |

Let $\u03f5>0$. Because $\mathcal{F}$ is equi-integrable, there is some $\delta >0$ such that when $A\in \mathrm{\Sigma}$ and $\mu (A)\le \delta $,

$$\underset{\alpha \in I}{sup}{\int}_{A}|{f}_{\alpha}|\mathit{d}\mu \le \u03f5,$$ |

which gives

$$|F({\chi}_{A})|=\underset{\alpha}{lim}\left|{\int}_{X}{f}_{\alpha}{\chi}_{A}\mathit{d}\mu \right|=\underset{\alpha}{lim}\left|{\int}_{A}{f}_{\alpha}\mathit{d}\mu \right|\le \underset{\alpha \in I}{sup}{\int}_{A}|{f}_{\alpha}|\mathit{d}\mu \le \u03f5.$$ |

By Theorem 1, this tells us that there is some $f\in {L}^{1}(\mu )$ for which

$$F(g)={\int}_{X}gf\mathit{d}\mu ,g\in {L}^{\mathrm{\infty}}(\mu ),$$ |

and hence $F=T(f)$. This shows that $\mathscr{H}\subset T({L}^{1}(\mu ))$, and

$${\int}_{X}{f}_{\alpha}g\mathit{d}\mu \to {\int}_{X}fg\mathit{d}\mu ,g\in {L}^{\mathrm{\infty}}(\mu )$$ |

tells us that ${f}_{\alpha}\to f$ in $\sigma ({L}^{1}(\mu ),{({L}^{1}(\mu ))}^{*})$, in other words ${T}^{-1}({F}_{\alpha})$ converges weakly to $T(F)$. Thus ${T}^{-1}:\mathscr{H}\to {L}^{1}(\mu )$ is continuous, where $\mathscr{H}$ has the subspace topology ${\tau}_{\mathscr{H}}$ inherited from ${({L}^{\mathrm{\infty}}(\mu ))}^{*}$ with the weak-* topology and ${L}^{1}(\mu )$ has the weak topology. $(\mathscr{H},{\tau}_{\mathscr{H}})$ is a compact topological space, so ${T}^{-1}(\mathscr{H})$ is a weakly compact subset of ${L}^{1}(\mu )$. But $\mathcal{F}\subset {T}^{-1}(\mathscr{H})$, which establishes that $\mathcal{F}$ is a relatively weakly compact subset of ${L}^{1}(\mu )$.

Suppose that $\mathcal{F}$ is a relatively compact subset of ${L}^{1}(\mu )$ with the weak topology and suppose by contradiction that $\mathcal{F}$ is not equi-integrable. Then by (2), there is some $\eta >0$ such that for all ${C}_{0}$ there is some $C\ge {C}_{0}$ such that

$$\underset{f\in \mathcal{F}}{sup}{\int}_{\{|f|>C\}}|f|\mathit{d}\mu >\eta ,$$ |

whence for each $n$ there is some ${f}_{n}\in \mathcal{F}$ with

$${\int}_{\{|{f}_{n}|>n\}}|{f}_{n}|\mathit{d}\mu \ge \eta ,$$ | (4) |

On the other hand, because $\mathcal{F}$ is relatively weakly compact, the Eberlein-Smulian theorem tells us that $\mathcal{F}$ is relatively weakly sequentially compact, and so there is a subsequence ${f}_{a(n)}$ of ${f}_{n}$ and some $f\in {L}^{1}(\mu )$ such that ${f}_{a(n)}$ converges weakly to $f$. For $A\in \mathrm{\Sigma}$, as ${\chi}_{A}\in {L}^{\mathrm{\infty}}(\mu )$ we have

$$\underset{n\to \mathrm{\infty}}{lim}{\int}_{A}{f}_{a(n)}\mathit{d}\mu ={\int}_{A}f\mathit{d}\mu ,$$ |

and thus Theorem 2 tells us that the collection $\{{f}_{a(n)}\}$ is equi-integrable, contradicting (4). Therefore, $\mathcal{F}$ is equi-integrable. ∎

###### Corollary 4.

Suppose that $(X,\mathrm{\Sigma},\mu )$ is a probability space. If $\{{f}_{n}\}\subset {L}^{1}(\mu )$ is bounded and equi-integrable, then there is a subsequence ${f}_{a(n)}$ of ${f}_{n}$ and some $f\in {L}^{1}(\mu )$ such that

$${\int}_{X}{f}_{a(n)}g\mathit{d}\mu \to {\int}_{X}fg\mathit{d}\mu ,g\in {L}^{\mathrm{\infty}}(\mu ).$$ |

###### Proof.

The Dunford-Pettis theorem tells us that $\{{f}_{n}\}$ is relatively weakly compact, so by the Eberlein-Smulian theorem, $\{{f}_{n}\}$ is relatively weakly sequentially compact, which yields the claim. ∎

## 5 Separable topological spaces

It is a fact that if $E$ is a separable topological vector space and $K$ is a compact subset of
$({E}^{*},\sigma ({E}^{*},E))$, then $K$ with the subspace topology inherited from
$({E}^{*},\sigma ({E}^{*},E))$ is metrizable. Using this and the Banach-Alaoglu theorem, if
$E$ is a separable normed space it follows that
$\{\lambda \in {E}^{*}:{\parallel \lambda \parallel}_{op}\le 1\}$
with the subspace topology inherited from
$(E,\sigma ({E}^{*},E))$ is compact and metrizable,
and hence is sequentially compact.^{7}^{7}
7
A second-countable ${T}_{1}$ space is compact if and only if it is
sequentially compact: Stephen Willard, General Topology, p. 125, 17G.
In particular, when $E$ is a separable normed space, a bounded sequence in ${E}^{*}$ has a weak-* convergent subsequence.

If $X$ is a separable metrizable space and $\mu $ is a $\sigma $-finite Borel measure on $X$, then
the Banach space ${L}^{p}(\mu )$ is separable for each $$.^{8}^{8}
8
René L. Schilling,
Measures, Integrals and Martingales, p. 270, Corollary 23.20.

###### Theorem 5.

Suppose that $X$ is a separable metrizable space and $\mu $ is a $\sigma $-finite Borel measure on $X$. If $\{{g}_{n}\}$ is a bounded subset of ${L}^{\mathrm{\infty}}(\mu )$, then there is a subsequence ${g}_{a(n)}$ of ${g}_{n}$ and some $g\in {L}^{\mathrm{\infty}}(\mu )$ such that

$${\int}_{X}f{g}_{a(n)}\mathit{d}\mu \to {\int}_{X}fg\mathit{d}\mu ,f\in {L}^{1}(\mu ).$$ |